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\begin{document}

\tableofcontents

\chapter{}

\subsection{Course Outline}

\subsubsection{Chapter 1}

Generalisation of E.M. radiation by fluctuating charges and currents (Hertz and Heaviside, 1880 - 90)

\subsubsection{Chapter 2}

The Lorentz symmetry of the Maxwell equations (1895).

This is \emph{not} the same as the symmetries of Newtonian physics; one of these formulations must be wrong.

\subsubsection{Chapter 3}

Covariant formulation of the generation of E.M. radiation. Describing the field from an arbitrary moving charge.

\subsubsection{Chapter 4}

To what extent are the maxwell equations a consequence of Lorentz invariance?

The action principle for electromagnetism.

\subsection{Course Content}

The Maxwell field theory is an example of a \emph{classical field theory}.

The key ideas are: Locality, causality and symmetry.

There is (almost) no quantum mechanics in the course.

However, by combining the classical theory of fields with quantum mechanics, we obtain \emph{quantum field theory}. The synthesis of these two fields occupied most of the period from 1930 - 70, and the resulting theoretical framework underpins all of modern fundamental physics.

Modern fundamental physics is addressed in the following 4 Masters level courses:

\begin{itemize}
\item RQFT
\item MQFT (Covering the same ideas as RQFT, but from a `better' theoretical perspective)
\item Standard Model
\item General Relativity
\end{itemize}

\subsection{Units}

The JH course in EM used SI units. A more convenient system of units is the ``Lorentz Heaviside'' units.

The constants $\epsilon_0$ and $\mu_0$ are scaled out from $\vec E$, $\vec B$, $\rho$ and $\vec J$. Specifically:

$$\vec E \rightarrow \frac{1}{\sqrt{\epsilon_0}} \vec E$$
$$\vec B \rightarrow \sqrt{mu_0} \vec B$$
$$\rho \rightarrow \sqrt{\epsilon_0} \rho$$
$$\vec J \rightarrow \sqrt{\epsilon_0} \vec J$$

This leaves $\rho \vec E \rightarrow \sqrt{\epsilon_0} \rho \frac{1}{\sqrt{\epsilon_0}} \vec E = \rho \vec E$, i.e. unchanged.

$$\vec J \times \vec B  = \sqrt{ \epsilon_0 \mu_0 } \vec J \times \vec B, \, c = \sqrt{ \epsilon_0 \mu_0 } $$

so the Lorentz force becomes

$$\vec F = \rho E + \frac{1}{c} \vec J \times \vec B$$

Now since $\vec J = \rho v$, and $v/c$ is dimensionless

$$[E] = [B]$$

i.e. $E$ and $B$ fields have the \emph{same} dimensions.

The maxwell equations (in vacuo) become
$$\vec \nabla \cdot \vec E = \rho$$
$$\vec \nabla \cdot \vec B = 0 \, \textrm{ (unchanged)}$$
$$\vec \nabla \times \vec E = \frac{1}{c} \frac{\partial \vec B}{\partial t}$$
$$\vec \nabla \times \vec B = \frac{1}{c} \vec J + \frac{1}{c} \frac{\partial \vec E}{\partial t}$$

All reference to $\mu_0$ and $\epsilon_0$ has disappeared. Two constants have been replaced by a single, more fundamental constant, $c$.

The stage is set for relativity, i.e. the disappearance of the ether.

There are also other systems of units, for example Gaussian units, in which $e \rightarrow 4 \pi e$ (in Heaviside Lorentz units).

\section{Conservation Laws}

We set out to show that the Maxwell equations imply conservation of charge, energy and momentum.

\subsection{Conservation of Charge}

To make a statement about \emph{sources} (i.e. charges and currents) we are obviously going to need the Maxwell equations which contain $\rho$ and $\J$ (the \emph{inhomogeneous} Maxwell equations)

$$\vec \nabla \cdot \vec E = \rho$$
$$\vec \nabla \times \vec B = \frac{1}{c} \vec J + \frac{1}{c} \frac{\partial \vec E}{\partial t}$$

To obtain a relation between $\rho$ and $\J$, substitute the electric equation into the magnetic one. First take the divergence of the magnetic equation

$$\div (\curl \B) = \frac{1}{c} \div \J + \frac{1}{c} \pdiff{}{t} ( \div \E )$$

where the left hand side is the divergence of a curl, and thus vanishes. Substitute for $\div \E$

$$0 = \frac{1}{c} \div \J + \frac{1}{c} \pdiff{}{t} \rho \Rightarrow \div \J = - \pdiff{\rho}{t}$$

which is a local equation relating $\J$ and $\rho$. To interpret this, integrate both sides over a volume $V$

$$\int \d V' \div \J(\r',t) = - \pdiff{}{t} \int \d V' \rho(\r',t)$$

apply the divergence theorem to the left hand side, and interpret the right hand integral as total enclosed charge $Q$

$$\oint_S \J(\r',t) \cdot \d \vec S = - \pdiff{Q}{t}$$

this is thus a statement of \emph{conservation of charge}; the charge within an enclosed volume can change only when some of it is carried out of the volume (across a bounding surface) as a current.

\subsection{Conservation of Energy}

An expression for the energy in $\E$ and $\B$ comes via the concept of \emph{work}, i.e.

$$W = \int_{\vec a}^{\vec b} \vec F \cdot \d \vec r$$

which is the work done in moving a particle from $\vec a$ to $\vec b$. Newtonian force $\vec F$ is connected to electromagnetism via the Lorentz force law

$$\vec f = \rho ( \E + \frac{\vec v}{c} \times \B )$$

which is an expression for $\vec f$, the force per unit volume. $\J = \rho \vec v$, i.e. $\vec v$ can be thought of as the velocity of a moving charge. We also make use of

$$\vec v = \diff{\r}{t} \Leftrightarrow \vec v \d t = \d \r$$

 Inserting $\vec f$ into the work integral, and using the above identity to make a change of integration variable from $\r$ to $t$
 
 $$w = \int_{t(\vec a)}^{t(\vec b)} \rho ( \E + \frac{\vec v}{c} \times \B ) \cdot \vec v \d t$$
 
 notice that this is an expression for work per unit volume $w$, rather than total work $W$. The term involving $\B$ vanishes, since $\vec v \cdot (\vec v \times \B) = \B \cdot (\vec v \times \vec v) = 0$; thus \emph{magnetic fields do no work}. We are left with the integral
 
$$w = \int_{t(\vec a)}^{t(\vec b)} \rho \vec v \cdot \E \d t = \int_{t(\vec a)}^{t(\vec b)} \J \cdot \E \d t$$

which we can dispense with by taking the time derivative of both sides

$$\diff{w}{t} = \J \cdot \E$$

Which is a local expression for the rate of working (power). We want to examine the energy in \emph{fields}, $\E$ and $\B$, so we eliminate $\J$ by using Ampere-Maxwell $\J = c \curl \B - \pdiff{\E}{t}$

$$\diff{w}{t} = \left(  c \curl \B - \pdiff{\E}{t} \right) \cdot \E$$

We can tidy up the time derivative by noting that $\pdiff{\E}{t} \cdot \E = \frac{1}{2} \pdiff{E^2}{t}$

$$\diff{w}{t} = c ( \curl \B ) \cdot \E - \pdiff{E^2}{t}$$

$$F_i = \oint_S d A_j T_{ij} - \frac{1}{c^2} \diff{}{t} \int \d V S_i$$

In a static situation

$$\diff{\vec p}{t} = 0$$
$$\vec F = \oint_S \tens T \cdot \d \vec A$$

This is the electromagnetic version of N3. In the absence of net forces on $V$, forces on the body are equal and opposite. $T_{ij} \d A_j$ is the force in the $i$th direction acting on a surface element $\d A_j$.

For dynamical systems, N2 means

$$\vec F = \diff{\vec p}{t}$$

$\vec p$ is the mechanical momentum of all charges within $V$, then

$$\diff{ \vec p}{t} + \diff{}{t} \int \d V \frac{1}{c^2} \vec S = \oint \tens T \cdot \d \vec A$$

If we write $\vec p = \int \d V \vec \rho$ where $\rho$ is the mechanical momentum \emph{density}, then locally

$$\pdiff{}{t} \left( \vec \rho + \frac{1}{c^2} \vec S \right) = \vec \nabla \cdot \tens T$$

Compare this with

$$\pdiff{}{t} \left( w + u \right) = - \vec \nabla \cdot \vec S$$

c.f. $-\vec \nabla \cdot \vec J$

Let us apply this to the example of plane waves in a vacuum, which satisfy the following equations

$$\vec E = \vec E_0 \phi( \vec k \cdot \vec r - \omega t )$$
$$\vec B = B_0 \vec n \times  \vec E_0 \phi( \vec k \cdot \vec r - \omega t )$$
$\omega^2 = c^2 k^2$, $\vec k = \frac{\omega}{c} \vec n$, $\vec n \cdot \vec E_0 = 0$

it is a useful exercise to check that these satisfy the Maxwell equations.

Let us calculate the \textbf{energy density}

\begin{eqnarray}
u = \frac{1}{2} \left( E^2 + B^2 \right) \\
&=& \frac{1}{2} \left( E_0^2 + \frac{1}{2} n^2 E_0^2 \right) \phi^2 = E_0^2 \phi^2
\end{eqnarray}

i.e. equal contributions from both $E$ and $B$ fields.

The \textbf{poynting vector} is given by

\begin{eqnarray}
\vec S &=& c \vec E \times \vec B \\
&=& c \vec E_0 \times ( \vec n \times \vec E_0 ) \phi^2 \\
&=& c E_0^2 \phi^2 \vec n
\end{eqnarray}

so the energy flow is $uc\vec n$, in the direction of the wave, with speed $c$.

The \textbf{momentum density} is

$$\vec \rho = \frac{1}{c^2} \vec S = \frac{1}{c} u \vec n $$

i.e. in the direction of the wave, and $u = c | \rho |$. Compare this with the relativistic energy-momentum relation

$$E^2 = m^2 c^4 + p^2 c^2$$

if we think ``quantum mechanically'' (i.e. in terms of particles), then the `particles' associated with electromagnetic waves are massless, i.e. \emph{photons}.

The \textbf{stress tensor} is

$$T_{ij} = \left( E_i^0 E_j^0 + (\vec n \times \vec E_0)_i (\vec n \times \vec E_0)_j - \delta_{ij} E_0^2 \right) \phi^2 $$

so $T_{ij} n_j = - n_i u$

\subsection{Angular Momentum}

Where there is momentum, we can define angular momentum

$$\vec l = \vec r \times \vec p = \frac{1}{c} \vec r \times ( \vec E \times \vec B)$$

Even \emph{static} fields can carry momentum and angular momentum. A nice illustration of this was given by Feynman

\subsubsection{Example (Charged Ring)}

Consider a disc of radius $a$, which is free to rotate about an axis through its centre. A charge of linear density $\lambda$ is on the edge of the disc. A magnetic field, confined to a cylindrical region of radius $a < b$, runs parallel to the rotation axis and intersects the disc.

If the field is switched off, what happens to the disc?

\textbf{Faraday's Law} (in Heaviside units) states

$$\oint_C \vec E \cdot \d \vec r = - \frac{1}{c} \int \d S \cdot \pdiff{\vec B}{t}$$

performing the surface integral gives

$$\oint_C \vec E \cdot \d \vec r = - \frac{\pi b^2}{c} \diff{B}{t}$$

The presence of an $E$ field exerts a torque on the disc.

\chapter{Radiation}

\section{Green's Functions}

A Green's function $G$ is the ``inverse'' of a differential operator $D$
$$D [f] = g \Rightarrow f = G \star g$$
where $\star$ represents convolution. c.f. the inverse of a matrix
$$A \vec w = \vec v \Rightarrow w = A^{-1} v$$

The method of Green's functions is now illustrated by applying it to two important differential equations, Poisson's equation and the wave equation.

\subsection{Poisson's Equation}

Poisson's equation is an inhomogeneous Laplace equation

$$\delsq \phi(\r) = - \rho(\r)$$

Consider substituting the convolution $G \star \rho$ in place of $\phi$

\begin{eqnarray*}
\delsq \phi(\r) &=& \delsq \left[ \int \d V' G(\r - \r') \rho(\r') \right] \\
&=& \int \d V' \delsq \left[ G(\r - \r') \right] \rho(\r') \\
\end{eqnarray*}

where we can pull the integral through the derivative, and $\nabla$ acts only on $\r$, not $\r'$. Now if $\delsq G(\r) = - \delta^3(\r)$, we have

\begin{eqnarray*}
\delsq \phi(\r) &=& - \int \d V' \delta^3(\r - \r') \rho(\r') \\
&=& - \rho(\r)
\end{eqnarray*}

where the delta function ``sifts'' the argument of the inhomogeneity $\rho$. So, we have reduced the problem of solving a differential equation to the problem of finding a $G(\r)$ such that $\delsq G(\r) = - \delta^3(\r)$. The solution $\phi$ is then given directly by the convolution $G \star \rho$, thus $G$ tells you how $\rho$ produces $\phi$.

The Green's function of a differential operator is not unique, since we can always \emph{add} any solution to the homogeneous equation i.e. a function $\chi$ such that $\delsq \chi = 0$, sending $\phi \rightarrow \phi + \chi$. In the (Newtonian?) language, $\phi$ is the ``particular integral'' and $\chi$ is the ``complementary function''.

To find $G$, remember that for a point charge located at the origin (i.e. a charge distribution $\rho(\r) = q \delta^3(\r)$), the potential is

$$\phi(\r) = \frac{1}{4 \pi} \frac{1}{| \r |}$$

if we compare this expression to the convolution of 

$$ \int \d V' \frac{1}{4 \pi} \frac{q}{| r |} \delta^3(\r - \r')$$

\subsection{Wave Equation}

The wave equation, with an inhomogeneity $\rho$, is

$$\delsq \phi - \frac{1}{c^2} \pdiffsq{}{t} \phi = - \rho$$

c.f. the Poisson equation, the wave equation depends on time also, hence we are seeking a Green's function of both space and time, such that

$$\delsq G(\r,t) - \frac{1}{c^2} \pdiffsq{}{t} G(\r,t) = - \delta^3(\r,t)$$

Assuming this fundamental property for the Green's function, we can check that the convolution over both space and time $G \star \rho$ satisfies the wave equation

\begin{eqnarray*}
\delsq \phi(\r) - \frac{1}{c^2} \pdiffsq{}{t} \phi &=& \delsq \left[ \int \d V' G(\r - \r',t-t') \rho(\r',t') \right] - \frac{1}{c^2} \pdiffsq{}{t} \left[ \int \d V' G(\r - \r',t-t') \rho(\r',t') \right]  \\
&=& \int \d V' \left( \delsq \left[ G(\r - \r',t-t') \right] - \pdiffsq{}{t} \left[ G(\r - \r',t-t') \right] \right) \rho(\r',t') 
\end{eqnarray*}

where, as in the case of the Poisson equation, we have commuted differentiation and integration, and use the fact that derivatives act on unprimed variables only. Now, using the fundamental property of the Green's function

\begin{eqnarray*}
\delsq \phi(\r) - \frac{1}{c^2} \pdiffsq{}{t} \phi &=& - \int \d V' \delta^3(\r - \r',t-t') \rho(\r',t') \\
&=& - \rho(\r,t)
\end{eqnarray*}

Now to find $G(\r,t)$ for the wave equation, note the spherical spatial symmetry of the differential operator (i.e. the Laplacian depends on $r$, not $\r$). This implies the same symmetry for $G$, i.e.

$$G = G(r,t)$$

The Laplacian, acting on a generic function $f$, can be re-written in spherical polars as

\begin{eqnarray*}
\delsq f &=& \frac{1}{r^2} \pdiff{}{t} \left( r^2 \pdiff{f}{r} \right) \\
&=& \pdiffsq{f}{r} + \frac{2}{r} \pdiff{f}{r} \\
&=& \frac{1}{r} \pdiffsq{}{r}(rf)
\end{eqnarray*}

Now the fundamental property of $G$ requires that $D[G] = 0$ away from $r = 0$, which implies an equation for $G$

$$\pdiffsq{}{r} (rG) - \frac{1}{c^2} \pdiffsq{}{t} (rG) = 0$$

(where we have applied the full wave equation operator to $G$, and multiplied through by $r$). This is just a 1 dimensional wave equation for the product $rG$, which has the solution (D'Alembert's solution) $rG = f(t \pm r/c)$, where $f$ is any function, so

$$G(\vec r, t) = \frac{ f( t \pm r/c) }{ r }$$

This describes a spherical wave, with amplitude decaying as $1 / r$. In fact, because the argument of $f$ can take on either sign, the Green's function describes two types of wave

$t \vec uparrow \Rightarrow r \vec uparrow$, is an \emph{outgoing} wave $f(t-r/c)$

$t \vec uparrow \Rightarrow r \downarrow$, is an \emph{incoming} wave $f(t-r/c)$

The advanced solution is \emph{not} causal; the wave appears before the source that creates it, which prompts the question ``what must the source be doing in the future in order to absorb the wave?''.

Often it is said that the advanced solution is ``unphysical'', and that the retarded solution is the only physical solution, but this is not really true (we will see why later in the course).

Finally, notice that both Poisson's equation and the wave equation are linear, and hence obey the superposition principle.

\section{Electromagnetic Potentials}

In electromagnetism it is in fact the potentials, not the fields, which are the fundamental objects. To solve Maxwell's equations with general sources $\rho(\vec r, t)$ and $\vec J(\vec r, t)$ is is much easier to work with the potentials $\phi$ and $\vec A$ than with the fields $\vec E$ and $\vec B$.

Absence of magnetic monopoles implies that $\vec B$ may be written as the curl of $\vec A$, the \emph{vector potential}.

$$ \div \vec B = 0 \Rightarrow \vec B = \curl \vec A$$

then

$$\curl \vec E = - \frac{1}{c} \pdiff{\vec B}{t} = - \frac{1}{c} \pdiff{}{t} ( \curl A )$$

by rearranging and pulling the curl through the time derivative

$$\Rightarrow \curl \left( \vec E + \frac{1}{c} \pdiff{\vec A}{t} \right) = 0$$

Helmholtz' theorem implies that the terms in brackets can be written as the gradient of a scalar $\phi$

$$\Rightarrow \vec E + \frac{1}{c} \pdiff{\vec A}{t} = - \grad \phi$$

so we have written both $E$ and $B$ in terms of potentials $\phi$ and $\vec A$

$$\vec E = - \grad \phi - \frac{1}{c} \pdiff{\vec A}{t}$$

$$\vec B = \curl \vec A$$

We can use the above results to rewrite the remaining `inhomogeneous' Maxwell equations (those involving sources) for electric fields

$$\div \vec E = \rho \Rightarrow \laplacian + \frac{1}{c} \pdiff{}{t} \div \vec A = - \rho$$

and whence, for magnetic fields

$$\curl B = \frac{1}{c} \left( \vec J + \pdiff{\vec E}{t} \right)$$
$$\Rightarrow \grad (\div \vec A) = \frac{1}{c} \vec J - \frac{1}{c} \pdiff{}{t} \grad \phi - \frac{1}{c^2} \pdiffsq{}{t} \vec A$$

and by employing the identity $\grad ( \div \vec A) - \laplacian \vec A$

$$\Rightarrow \laplacian \vec A - \frac{1}{c^2} \pdiffsq{}{t} \vec A - \grad \left( \div \vec A + \frac{1}{c} \pdiff{\phi}{t} \right) = - \frac{1}{c} \vec J$$

We can rewrite the equation for $\phi$ to resemble the equation for $\vec A$, by adding and subtracting a term $\frac{1}{c^2} \pdiffsq{\phi}{t}$, to obtain

$$\laplacian \phi - \frac{1}{c^2} \pdiffsq{\phi}{t} + \frac{1}{c} \pdiff{}{t} \left( \div \vec A + \frac{1}{c} \pdiff{\phi}{t} \right)= - \rho$$

This is a very messy expression, but we can apply a \emph{gauge transformation} to simplify things. We can amend $\vec A$ according to

$$\vec A \rightarrow \vec A + \grad \chi$$

which leaves $\vec B = \curl \vec A$ unchanged, for any scalar function $\chi$. This transformation alone \emph{would} change $\vec E$, and so in order to cancel this we must also transform $\phi$ according to

$$\phi \rightarrow \phi - \frac{1}{c} \pdiff{\chi}{t} $$

which leaves $\vec E$ unchanged also. We can use this ``gauge freedom'' to put restrictions on $\vec A$ and $\phi$.

We choose to work in ``Lorenz gauge'', such that

$$\div \vec A + \frac{1}{c} \pdiff{\phi}{t} = 0$$

To find the necessary transformation, suppose

$\div A + \frac{1}{c} \pdiff{\phi}{t} = \psi \neq 0$

then we should choose $\chi$ such that

$$\div \vec A' + \frac{1}{c} \pdiff{\phi'}{t} = \psi + \laplacian \chi - \frac{1}{c^2} \pdiffsq{\chi}{t} = 0$$

where primes denote the transformed potentials, which implies

$$\laplacian \chi - \frac{1}{c^2} \pdiffsq{\chi}{t} = - \psi$$

which is a wave equation (i.e. it has unique solutions for some given boundary conditions), hence we can \emph{always} find a Lorenz gauge. Applying this transformation to the equations for electric and magnetic potentials, we obtain

$$\laplacian \phi - \frac{1}{c^2} \pdiffsq{\phi}{t} = - \rho$$

$$\laplacian \vec A - \frac{1}{c^2} \pdiffsq{\vec A}{t} = \frac{1}{c} \vec J$$

i.e. both $\phi$ and $\vec A$ satisfy inhomogeneous wave equations! Compare these with the analogous Poisson equations in electrostatics and magnetostatics

$$\laplacian \phi = - \rho$$

$$\laplacian \vec A = - \frac{1}{c^2}$$

The solutions to these equations, using the retarded Green's functions, are

$$\phi(\vec r,t) = \frac{1}{4 \pi} \int \d V' \frac{ \rho( \vec r', \tau ) }{ | \vec r - \vec r' | }$$
$$\vec A (\vec r,t) = \frac{1}{4 \pi c} \int \d V' \frac{ \vec J( \vec r', \tau ) }{ | \vec r - \vec r' | }$$

where $\tau$ is the \emph{retarded time}, defined by

$$\tau = t - | \vec r - \vec r' | / c$$

Compare these solutions to \emph{electrodynamics} with the solutions to \emph{electrostatics}

$$\phi( \vec r ) = \frac{1}{4 \pi} \int \d V' \frac{ \rho( \vec r' ) }{ | \vec r - \vec r' | }$$
$$\vec A(\vec r) = \frac{1}{4 \pi c} \int \d V' \frac{ \vec J(\vec r') }{ | \vec r - \vec r' | }$$

in which there is no time dependence.

\subsection{Physical Implications}

\begin{figure}[htb]
\begin{center}
	\includegraphics[width=0.8\textwidth]{dynamicalsolution}
	\caption{Physical interpretation of solution to inhomogeneous Maxwell equations}
	\label{dynamicalsolution}
\end{center}
\end{figure}

Physically, these equations describe the potentials experienced by an observer at $\vec r$, originating from a source at $\vec r'$. The observer and source are separated by a distance $| \vec r - \vec r' |$. Information about the configuration of the sources has to travel from source to observer, and it does so at the speed of light, hence it takes a time

$$\Delta t = | \vec r - \vec r' | / c$$

to reach the observer. The observer `sees' the source as it was at the \emph{retarded time}

$$\tau = t - \Delta t$$

Other than this, the solutions are just the same as in statics!

This reminds us of astronomical observations, in which objects observed through a telescope are images of ancient galaxies. Thus we may be tempted to think that this retardation effect is only important when the distances or times involved are very large, however we will see that this is of vital importance at \emph{all} time scales.

\subsection{Causality}

Notice that we chose to use the solution for the \emph{retarded}, rather than the \emph{advanced} time. This means that effects at $\vec r$ happen \emph{after} the `wiggles' (changes to the source!) that caused them, \emph{not} before.

The causal relationship is usually illustrated with a ``light cone''.

\begin{figure}[htb]
\begin{center}
	\includegraphics[width=0.8\textwidth]{lightcone}
	\caption{Light cone}
	\label{lightcone}
\end{center}
\end{figure}

Note that in electromagnetism, there is \emph{no action at a distance}; effects cannot travel instantaneously, but only at the speed of light.

This embodies the physical principle of \emph{locality}, which is a basic philosophical assumption present in all physical laws.

Notice that quantum mechanics is fact a local theory. Confusion over this (e.g. entanglement paradox) arises from the \emph{logical} implications of QM, rather than the \emph{physical} implications.

\subsection{Jefimenko Equations}

The solutions that we found are for the potentials, which may still seem `unphysical'. To find equations for the fields, we need to differentiate with respect to $t$ and $\vec r$ ($t'$ and $\vec r'$ stay fixed).

By applying the product rule to $\phi$

$$\grad \phi =  \frac{1}{4 \pi} \int \d V' \left\{ \frac{1}{| \vec r - \vec r' |} \grad \rho( \vec r', \tau ) + \rho( \vec r', \tau ) \grad \left( \frac{1}{| \vec r - \vec r' |} \right) \right\}$$

now

$$\grad \rho( \vec r', \tau ) = \pdiff{\rho}{t} \grad \tau = - \frac{1}{c} \pdiff{\rho}{t} \grad ( | \vec r - \vec r' | )$$

so

$$\grad \phi = \frac{1}{4\pi} \int \d V' \left( \frac{1}{c} \pdiff{\rho}{t} + \frac{\rho}{|\vec r - \vec r'|} \right) \frac{\vec r - \vec r'}{|\vec r - \vec r'|^2}$$

inserting the solution for $\grad \phi$ to the expression for $\vec E$ gives

$$\vec E = \frac{1}{4\pi} \int \d V' \left\{ \left( \frac{1}{c} \pdiff{\rho}{t} + \frac{\rho}{|\vec r - \vec r'|} \right) \frac{\vec r - \vec r'}{|\vec r - \vec r'|^2} - \frac{1}{c^2} \pdiff{\vec J}{t} \frac{1}{|\vec r - \vec r'|} \right\} $$

Similarly, for the $\vec B$ field

$$\curl \vec A = \frac{1}{4\pi c} \int \d V' \left\{ \frac{1}{|\vec r - \vec r'|} \curl \vec J(\vec r', \tau) - \vec J(\vec r', \tau) \times \grad \left( \frac{1}{|\vec r - \vec r'|} \right) \right\}$$

now

$$\curl \vec J (\vec r', \tau) = - \pdiff{\vec J}{t} \times \grad \tau = - \frac{1}{c} \pdiff{\vec J}{t} \times \grad ( \vec r - \vec r' )$$

so

$$\vec B = \frac{1}{4\pi c} \int \d V' \left\{ \frac{1}{c} \pdiff{\vec J}{t} + \frac{\vec J}{| \vec r - \vec r' |} \right\} \times \frac{ \vec r - \vec r' }{| \vec r - \vec r' |}$$

These equations for the fields $\vec E$ and $\vec B$ are called the ``Jefimenko equations". They show explicitly that there is \emph{no action at a distance} in electromagnetism (since both $\vec E$ and $\vec B$ depend on $\tau$, not $t$). Also note that these equations are gauge invariant, as indeed they must be.

\section{Oscillating Sources}

Consider a time-dependent charge $\rho(\vec r,t)$ and current $\vec J(\vec r,t)$. We can assume a particular functional form for the time dependence of

$$\rho(\vec r,t) = \rho(\vec r) e^{i \omega t}$$
$$\vec J(\vec r,t) = \vec J(\vec r) e^{i \omega t}$$

without loss of generality. This is because the Maxwell equations are linear, and so the solutions to these oscillating sources will obey the superposition principle, hence we can superimpose modes to build up any desired time dependence (note that this technique will turn out like a ``manual'' Fourier transform).

Recall the retarded solution for scalar potential in Lorentz gauge

$$\phi(\vec r,t) = \phi(\vec r) e^{i \omega t} = \frac{1}{4 \pi} \int \d V' \frac{ \rho(\vec r', \tau) }{ | \vec r - \vec r' | }$$

where the time within the integral is \emph{retarded time}, $\tau = t - |\r - \r'| / c$. We can separate the $\tau$ dependence within the integral

\begin{eqnarray*}
\phi(\vec r) e^{i \omega t} &=& \frac{1}{4 \pi} \int \d V' \frac{ \rho(\vec r') }{ | \vec r - \vec r' | } e^{i \omega (t - |\r - \r'| / c)} \\
\Rightarrow \phi(\r) &=& \frac{1}{4 \pi} \int \d V' \frac{ \rho (\r') }{| \r - \r' |} e^{- i k |\r - \r'| }
\end{eqnarray*}

Where we define $k = \omega / c$. A similar procedure can be applied to determine the spatial dependence of the vector potential $\A ( \r )$

$$\vec A (\vec r) = \frac{1}{4 \pi c} \int \d V' \frac{ \vec J (\vec r') }{ | \vec r - \vec r' | } e^{-i k | \vec r - \vec r' | }$$

In general, the fields are then given by

$$\vec B ( \vec r ) = \curl \vec A ( \vec r )$$
$$\vec E ( \vec r ) = - \grad \phi - \frac{1}{c} \pdiff{\vec A}{t}$$

however if we are concerned with radiation \emph{outside} the region of charge/current, where $\J=0$ i.e. far away from the sources, we can use

$$\curl \B(\r,t) = \frac{1}{c} \pdiff{}{t} \E(\r,t) = \frac{1}{c} \pdiff{}{t} \left( \E(\r) e^{i \omega t} \right) = \frac{i \omega}{c} \E(\r,t)$$

to obtain the $\vec B$ field, since the time derivative acts on the complex exponential only. We can then invert the above to obtain the $\E$ field

$$ \Rightarrow \E(\r,t) = \frac{1}{ik} \curl \vec B$$

So, in practice it is sufficient to compute $\vec A$ in order to determine both electric and magnetic fields, and we can ignore $\phi$. If (for some reason) $\phi$ is needed, we can use the Lorentz condition, i.e.

$$\div \vec A + \frac{1}{c} \pdiff{\phi}{t} = 0 \Rightarrow \phi = \frac{1}{ik} \div \vec A$$

Now, assume that $\vec J$ is localised in a region of approximate size $d$

\begin{figure}[htb]
\begin{center}
	\includegraphics[width=0.8\textwidth]{radiation}
	\caption{Radiation from an oscillating source}
	\label{radiation}
\end{center}
\end{figure}

Assume that

$$\lambda = \frac{2 \pi}{k} = \frac{2 \pi c}{\omega} \>\> d$$

i.e. that the wavelength of the radiation is much greater than the size of the current distribution.

There are 3 different length scales in this problem, $d$, $\lambda$ and $r$.

We thus have 3 different regions to consider:

Near (static) zone $d << r << \lambda$

Intermediate (induction) zone $d << r \sim \lambda$

Far (radiation) zone $d << \lambda << r$

In all these cases, $r >> d$ and $\lambda >> d$

$$\lambda >> d \Rightarrow \frac{\omega d}{c}$$

and $\omega d \sim v$ (the velocity of the charges) so $v / c << 1$, i.e. the charges are moving \emph{non-relativistically}.

Now, working in the far zone, we want to evaluate the integral for the vector potential, with the assumed harmonic time dependence.

$$\vec A (\vec r) = \frac{1}{4 \pi c} \int \d V' \frac{ \vec J (\vec r') }{ | \vec r - \vec r' | } e^{-i k | \vec r - \vec r' | }$$

The problematic part of this integral is the distance from the charge/current distribution $|\r - \r'|$. To evaluate the integral we make a series of expansions of  $|\r - \r'|$, in powers of $\r$, and apply our assumptions about the far-field regime ( $r' = d \ll r$ ) and the wavelength of the radiation ( $k = \omega / c \Rightarrow kr' \sim v / c \ll 1$ ) to remove higher-order terms.

Taylor expanding the entire integrand ab-initio would be complicated, fortunately we can expand each instance of $|\r - \r'|$ separately, then combine the expansions and drop higher order terms afterwards, provided that we keep track of powers of $\r$, $\r'$ and $k$. When writing $\mathcal{O}$ to keep track of powers, remember that the dimensions must match across the equation, including those terms within the $\mathcal{O}$.

First we re-write $| \r - \r' |$, and apply the binomial expansion

\begin{eqnarray*}
| \vec r - \vec r' | &=& \left( r^2 - 2 \vec r \cdot \vec r' + r'^2 \right)^{1/2} \\
&=& r \left( 1 - 2 \frac{ \hat \r \cdot \r' }{r} + \frac{r'^2}{r^2} \right)^{1/2} \\
&=& r \left[ 1 - \frac{ \hat \r \cdot \r' }{r} + \frac{1}{2} \frac{r'^2}{r^2} + \mathcal{O} \left( \frac{d^2}{r^2} \right) \right] \\
&=& r - \hat \r \cdot \r' + \frac{1}{2} \frac{r'^2}{r} + \mathcal{O} \left( \frac{d^2}{r} \right) \\
&=& r - \hat \r \cdot \r' + \mathcal{O} \left( \frac{d^2}{r} \right) \\
\end{eqnarray*}

Now $| \r - \r' |$ appears in both the denominator of the integrand, and in the exponential. We simply plug the above expansion into a binomial expansion for the denominator

\begin{eqnarray*}
\left[ r - \hat \r \cdot \r' + \mathcal{O} \left( \frac{d^2}{r} \right) \right]^{-1} &=& \frac{1}{r} \left[ 1 - \frac{ \hat \r \cdot \r' }{r} + \mathcal{O} \left( \frac{d^2}{r^2} \right) \right]^{-1} \\
&=& \frac{1}{r} \left[ 1 + \frac{ \hat \r \cdot \r' }{ r } + \order{ \frac{d^2}{r^2} } \right] \\
&=& \frac{1}{r} + \frac{ \hat \r \cdot \r' }{ r^2 } + \order{ \frac{d^2}{r^2} }
\end{eqnarray*}

and into the well-known power series expansion for the exponential

\begin{eqnarray*}
\exp \left\{ -ik \left[  r - \hat \r \cdot \r' + \mathcal{O} \left( \frac{d^2}{r} \right)  \right] \right\} &=& 
e^{ -ikr } \exp \left\{ ik \hat \r \cdot \r' + \order{ \frac{kd^2}{r} } \right\} \\
&=& e^{ -ikr } \left[ 1 + ik \hat \r \cdot \r' + \order{ \frac{kd^2}{r} } \right] \\
\end{eqnarray*}

The integrand involves the product of the above two expansions.

$$ e^{ -ikr } \left[ 1 + ik \hat \r \cdot \r' + \order{ \frac{kd^2}{r} } \right] \left[ \frac{1}{r} + \frac{ \hat \r \cdot \r' }{ r^2 } + \order{ \frac{d^2}{r^2} } \right] $$



\subsection{Electric Dipole Radiation}

Consider the leading order term

$$ \frac{ e^{-ikr} }{ 4 \pi c r } \int \d V' \J ( \r' ) $$

this is called the `electric dipole' term. To see why, recall the charge conservation equation

$$ \pdiff{ \rho }{ t } = - \div \J$$

Now consider a surface at infinity, beyond the compact charge-containing region. Clearly the current flux through this surface is zero

$$ \int_{S_\infty} \J \cdot \d \vec A = 0$$

indeed, any integrand with a factor of $\J$ must vanish, as is the case with the following tensor integral

\begin{eqnarray*}
0 &=& \int_{S \rightarrow \infty} r'_i J'_j \d A'_j \\
&=& \int_{V \rightarrow \infty} \partial'_j ( r'_i J'_j ) \d V' \\
&=& \int_{V \rightarrow \infty} \left\{ r'_i \partial'_j J'_j + J'_j \partial'_j r_i \right\} \d V' \\
&=& \int_{V \rightarrow \infty} \left\{ r'_i \left( - \pdiff{ \rho }{ t} \right) + J'_j \delta_{ij} \right\} \d V' \\
&\Rightarrow& \int_{V \rightarrow \infty} \d V' r'_i \pdiff{ \rho }{ t} = \int_{V \rightarrow \infty} \d V' J'_i
\end{eqnarray*}

now using the harmonic time dependence $\rho = \rho(\r) e^{i \omega t}$, $\J = \J(\r) e^{i \omega t}$

\begin{eqnarray*}
\int_{V \rightarrow \infty} \d V' r'_i \pdiff{}{t} \left(  \rho(\r') e^{i \omega t} \right) &=& \int_{V \rightarrow \infty} \d V' J'_i(\r')e^{i \omega t} \\
\Rightarrow i \omega e^{i \omega t}  \int_{V \rightarrow \infty} \d V' r'_i \rho(\r') &=& e^{i \omega t}  \int_{V \rightarrow \infty} \d V' J'_i(\r') \\
\Rightarrow i \omega \int_{V \rightarrow \infty} \d V' \r' \rho(\r') &=& \int_{V \rightarrow \infty} \d V' \J'(\r') \\
\end{eqnarray*}

finally, if we recall the definition of the dipole moment for a pair of point charges

$$\vec p = q \vec d$$

the natural extension of this to a continuous charge distribution is 

$$\vec p = \int_V \d V \rho(\r) \r$$

so we identify the above integral on the LHS with the electric dipole moment of the charge distribution

$$i \omega \vec p = \int_{V \rightarrow \infty} \d V' \J'(\r')$$

thus inserting this into the leading order term for the vector potential

$$\A (\r) = \frac{e^{-ikr}}{r} \frac{ik}{4 \pi} \vec p$$

\subsection{Fields from Electric Dipole Radiation}

To obtain the electric and magnetic fields, we need to take curls of the vector potential.

\begin{eqnarray*}
\B(\r) &=& \curl \A(\r) = \curl \frac{e^{-ikr}}{r} \frac{ik}{4 \pi} \vec p \\
&=& \frac{e^{-ikr}}{r} \frac{ik}{4 \pi} \cancel{ \curl \vec p } + \frac{ik}{4 \pi} \grad \left( \frac{e^{-ikr}}{r} \right) \times \vec p \\
&=&  \frac{ik}{4 \pi} \left[ \frac{1}{r} \grad \left( e^{-ikr} \right) + e^{-ikr} \grad \left( \frac{1}{r} \right) \right] \times \vec p \\
&=&  \frac{ik}{4 \pi} \left[ \frac{1}{r} \left( -ik \hat \r e^{-ikr} \right) + e^{-ikr} \left( - \frac{\hat \r}{r^2} \right) \right] \times \vec p \\
&=&  - \frac{ik}{4 \pi r} e^{-ikr} \left[ ik + \frac{1}{r} \right] \hat \r \times \vec p \\
\end{eqnarray*}

where $\curl \vec p$ vanishes as $\vec p$ is a constant vector.

Recalling the relation for the electric field

$$\E ( \r ) = \frac{1}{ik} \curl \B(\r)$$

in order to obtain the electric field, we must perform another curl, which could be quite messy. To simplify the calculation, note that

$$\B(\r) = f(r) \vec g (\r)$$

i.e. $\B$ is the product of a scalar function of a scalar argument, and a vector function of a vector argument. The curl of this generic product is

\begin{eqnarray*}
\curl \left( f(r) \vec g (\r) \right) &=& \grad \left( f(r) \right) \times \vec g (\vec r) + f(r) \curl \vec g(\r) \\
&=& \diff{f(r)}{r} \grad ( r ) \times \vec g (\vec r) + f(r) \curl \vec g(\r) \\
&=& \diff{f(r)}{r} \hat \r \times \vec g (\vec r) + f(r) \curl \vec g(\r) \\
\end{eqnarray*}

where we have used the chain rule on $f(r)$. Identifying $f(r)$

\begin{eqnarray*}
f(r) &=& - \frac{ik}{4 \pi} \frac{e^{-ikr}}{r} \left( ik + \frac{1}{r} \right) \\
\diff{}{r} f(r) &=& \left[ -ik - \frac{1}{r} - \frac{1}{r \left( ik + 1/r \right)} \right] f(r) \\
&=& \frac{ik}{4 \pi} \frac{e^{-ikr}}{r} \left[ ik  \left( ik + \frac{1}{r} \right) + \frac{ \left( ik + \frac{1}{r} \right) }{ r } + \frac{1}{r} \right] \\
&=& \frac{ik}{4 \pi} \frac{e^{-ikr}}{r^2} \left[ -k^2 r + ikr + ik + \frac{1}{r} + 1\right] \\
&=& \frac{ik}{4 \pi} \frac{e^{-ikr}}{r^3} \left[ (ik - k^2) r^2 + (ik + 1)r + 1\right] \\
\end{eqnarray*}

and $\vec g(\r)$

\begin{eqnarray*}
\vec g(\r) &=&  \hat \r \times \vec p \\
\curl \vec g(\r) &=& \curl (\hat r \times \vec p) \\
\Rightarrow \left[ \curl \vec g(\r) \right]_i &=& \epsilon_{ijk} \partial_j \epsilon_{klm} \hat r_l p_m \\
&=& (\delta_{il} \delta_{jm} - \delta_{im} \delta_{jl} ) \partial_j ( \hat r_l p_m ) \\
&=& \partial_j ( \hat r_i p_j ) - \partial j ( \hat r_j p_i ) \\
&=& p_j \partial_j \hat r_i + \hat r_i \partial_j p_j - p_i \partial_j \hat r_j - \hat r_j \partial_j p_i \\
&=& p_j \partial_j \hat r_i - p_i \partial_j \hat r_j
\end{eqnarray*}

now since $\hat r_i = r_i (r_j r_j)^{-1/2}$

$$\partial_j \hat r_i = \frac{1}{r} \partial_j r_i + r_i \partial_j (r_k r_k)^{-1/2} = \frac{\delta_{ij}}{r} - \frac{ r_i r_j }{ r^3 }$$

$$\partial_j \hat r_j = \frac{1}{r} \partial_j r_j + r_j \partial_j (r_k r_k)^{-1/2} = \frac{\delta_{jj}}{r} - \frac{ r_j r_j }{ r^3 } = \frac{3}{r} - \frac{1}{r} = \frac{2}{r}$$

so we have

\begin{eqnarray*}
\left[ \curl \vec g(\r) \right]_i &=& p_j \left( \frac{\delta_{ij}}{r} - \frac{ r_i r_j }{ r^3 } \right) - p_i \frac{2}{r} \\
&=& p_i \frac{1}{r} - \frac{ r_i \r \cdot \vec p }{ r^3 } - p_i \frac{2}{r} \\
\Rightarrow \curl \vec g(\r) &=& - \frac{1}{r} \left( \vec p + \r \frac{\r \cdot \vec p}{r^2} \right)
\end{eqnarray*}

\chapter{Relativity}

\section{Lorentz Transformations}

Fizeau (1851) used an interferometer to measure the velocity of light in a moving medium (a pipe with flowing water). He discovered the following empirical formula describing the speed of light in the medium of refractive index $n$, moving at velocity $v$

$$ \frac{c}{n} + v ( 1 - \frac{1}{n^2} )$$

the extra term (where $(1-1/n^2)$ is called the ``Fresnel drag coefficient'') was thought to describe the water `dragging' the light along.

Lorentz attempted to explain this effect in terms of coordinate transformations. He began by considering the propagation of light described by the wave equation

$$\left( \delsq - \frac{1}{c^2} \pdiffsq{}{t} \right) \phi = 0$$

due to Maxwell. Lorentz considered the Galilean transform

$$\vec x' = \vec x - \vec v t, \, \, t' = t$$

which describes the motion in a frame moving with velocity $\vec v$. Differential operators in this new frame transform according to

\begin{eqnarray*}
\pdiff{}{x_i} &=& \pdiff{x'_j}{x_i} \pdiff{}{x'_j} + \pdiff{t'}{x_i} \pdiff{}{t'} \\
&=& \pdiff{}{x_i} ( x_j - v_j t ) \pdiff{}{x'_j} + \pdiff{}{x_i} ( t ) \pdiff{}{t'} \\
&=& \pdiff{x_j}{x_i} \pdiff{}{x'_j} = \delta_{ij} \pdiff{}{x'_j} = \pdiff{}{x'_i}
\end{eqnarray*}

\begin{eqnarray*}
\pdiff{}{t} &=& \pdiff{x'_j}{t} \pdiff{}{x'_j} + \pdiff{t'}{t} \pdiff{}{t'} \\
&=& \pdiff{}{t} (x_j - v_j t) \pdiff{}{x_j} + \pdiff{}{t} ( t ) \pdiff{}{t'} \\
&=& \pdiff{}{t'} - v_j \pdiff{}{x_j}
\end{eqnarray*}

where we have applied the chain rule, and then substituted for primed quantities using the Galilean transformation. The gradient operator is unchanged, however the time derivative gains an extra term $-v_i \pdiff{}{x_i}$. The equation for $\phi$ in the moving frame is then

$$ \left\{ \delsqprime - \frac{1}{c^2} \left( \pdiff{}{t'} - \vec v \cdot \nabla' \right)^2 \right\} \phi = 0 $$

and squaring out the operator in parentheses

$$\left( \delsqprime - \frac{1}{c^2} \pdiffsq{}{t'} + \frac{2 \vec v \cdot \nabla'}{c^2} - \frac{(\vec v \cdot \nabla')^2}{c^2} \right) \phi = 0$$

Notice that this is \emph{no longer} a wave equation. The light should continue to propagate as a wave, even in the moving medium, and so Lorentz contemplated the transformation that would be necessary to recover a wave equation in the moving frame. He noticed (1895) that if we transformed the time coordinate according to

$$t' = t - \frac{ \vec v \cdot \vec x }{ c^2 }$$

to obtain a quantity referred to as the `local time', then the transformed gradient operator could be made to resemble the transformed time operator

\begin{eqnarray*}
\pdiff{}{x_i} &=& \pdiff{x'_j}{x_i} \pdiff{}{x'_j} + \pdiff{t'}{x_i} \pdiff{}{t'} \\
&=& \pdiff{}{x_i} ( x_j - v_j t ) \pdiff{}{x'_j} + \pdiff{}{x_i} ( t - v_k x_k / c^2 ) \pdiff{}{t'} \\
&=& \pdiff{x_j}{x_i} \pdiff{}{x'_j} - \frac{ v_k }{c^2} \pdiff{x_k}{x_i} \pdiff{}{t'} \\
&=& \delta_{ij} \pdiff{}{x'_j} - \frac{ v_k }{c^2} \delta_{ki} \pdiff{}{t'} = \pdiff{}{x'_i} - \frac{ v_i }{c^2} \pdiff{}{t'}
\end{eqnarray*}

and the resulting differential equation in the new frame would be

$$\left\{ \left( \nabla' - \frac{\vec v}{c^2} \pdiff{}{t'} \right)^2 -  \frac{1}{c^2} \left( \pdiff{}{t'} - \vec v \cdot \nabla' \right)^2 \right\} \phi = 0$$

and again, squaring the operators out, we obtain

$$\left( \delsqprime - \frac{1}{c^2} \pdiffsq{}{t'} - \frac{(\vec v \cdot \nabla')^2}{c^2} + \frac{v^2}{c^4} \pdiffsq{}{t'} \right) \phi = 0$$

which is \emph{almost} a wave equation; the first two terms on the left constitute a wave equation for waves of velocity $c$, and the others terms are $\mathcal O (v^2 / c^2)$. Successive attempts to obtain the exact wave equation in the transformed frame resulted in higher-order remainder terms, but it was Lorentz who obtained the correct transformation. He did so by requiring that the transformation be \emph{invertible}, i.e.

$$(\vec x, t) \rightarrow_{\vec v} (\vec x', t' ) \rightarrow_{- \vec v} (\vec x, t)$$

which is of course a perfectly sensible physical requirement, and is satisfied for Galilean transformations.

With no particular motivation, consider the rescaling of $\vec x $ by a factor $\gamma$, and $t$ by a factor $\bar \gamma$. The transformed coordinates would thus be

$$\vec x' = \gamma ( \vec x - \vec v t )$$
$$t' = \bar \gamma (t - \vec v \cdot \vec x / c^2)$$

The inverse of the time transformation would be

$$t = \bar \gamma (t' + \vec v \cdot \vec x' / c^2)$$

Inserting the transformed time $t'$ into the inverse transformation above should recover $t = t$

$$t = \bar \gamma ( \bar \gamma (t - \vec v \cdot \vec x / c^2) + \vec v \cdot \vec x' / c^2 )$$

we can eliminate $\vec v \cdot \vec x'$ from the above by inserting the dot product of $\vec v$ with the transformed vector $\vec x'$

$$\vec v \cdot \vec x' = \vec v \cdot \gamma ( \vec x - \vec v t ) = \gamma ( \vec v \cdot \vec x - v^2 t )$$

we obtain

\begin{eqnarray*}
t &=& \bar \gamma ( \bar \gamma (t - \vec v \cdot \vec x / c^2) + \gamma ( \vec v \cdot \vec x - v^2 t ) / c^2 ) \\
&=& \bar \gamma^2 ( t - \vec v \cdot \vec x / c^2 ) + \bar \gamma \gamma ( \vec v \cdot \vec x - v^2 t ) / c^2 \\
\end{eqnarray*}

now if $\gamma = \bar \gamma$, the above equation reduces to

$$t = \gamma^2 \left( 1 - \frac{v^2}{c^2} \right) t$$

hence, in order to recover the desired identity $t = t$, we can identify 

$$\gamma = \frac{1}{ \sqrt{ 1 - \frac{v^2}{c^2} } }$$

i.e. the venerable Lorentz factor.

A similar condition exists for the transformed position vector $\vec x$

\begin{eqnarray*}
\vec x &=& \gamma ( \vec x' + \vec v t' ) \\
&=& \gamma ( \gamma ( \vec x - \vec v t ) + \vec v \bar \gamma (t - \vec v \cdot \vec x / c^2 ) \\
&=& \gamma^2 ( \vec x - \vec v t ) + \bar \gamma \gamma \vec v ( t - \vec v \cdot \vec x / c^2 )
\end{eqnarray*}

and again, if $\gamma = \bar \gamma$ we have

$$\vec x = \gamma^2 ( \vec x - \vec v \cdot \vec x / c^2 )$$

taking the dot product with $\vec v$

$$\vec v \cdot \vec x = \gamma^2 ( \vec v \cdot \vec x - v^2 (\vec v \cdot \vec x) / c^2) = \gamma^2 (\vec v \cdot \vec x) \left( 1 - \frac{v^2}{c^2} \right) $$

and since $\vec v$ is arbitrary, in order to recover $\vec x = \vec x$ we require that $\gamma$ be the same Lorentz factor.

Notice that if we write $\vec x$ as the sum of parallel and perpendicular components

$$\vec x = \vec x_\perp + \vec x_\parallel$$

and choose our coordinates (via a rotation) such that $\vec x_\perp \cdot \vec v = 0$, then the perpendicular transformation is trivially unchanged, and the coordinates transform according to

$$\vec x'_\perp = \vec x_\perp$$
$$\vec x'_\parallel = \gamma ( \vec x_\parallel - \vec v t )$$
$$t' = \gamma ( t - \vec v \cdot \vec x / c^2 )$$

which are of course the exactly invertible \emph{Lorentz transformations}.

Returning to the wave equation, let us investigate how the differential operators behave under Lorentz transformation. Recall that, with the local time $t' = t - \vec v \cdot \vec x / c^2$ the gradient operator transforms according to

$$\pdiff{}{x_i} = \pdiff{}{x'_i} - \frac{ v_i }{c^2} \pdiff{}{t'}$$

We are free to decompose the gradient operator into parallel and perpendicular components

$$\nabla = \nabla_\perp + \nabla_\parallel$$

and if we choose the parallel component along the direction of $\vec v$, then the components of the gradient operator transform according to

$$\nabla_\perp = \nabla'_\perp$$
$$\nabla_\parallel = \gamma \left( \nabla'_\parallel - \frac{ 1 }{c^2} \vec v \pdiff{}{t'} \right)$$

since $\vec v \cdot \vec x$ vanishes along the perpendicular directions. The complete transformed gradient is then

$$\nabla = \nabla'_\perp + \gamma \left( \nabla'_\parallel - \frac{ 1 }{c^2} \vec v \pdiff{}{t'} \right)$$

The time operator transforms according to

$$\pdiff{}{t} = \gamma \left( \pdiff{}{t'} -  \vec v \cdot \nabla'_\parallel \right)$$

where the perpendicular component is absent for the same reason. Now we can substitute these transformed operators into the wave equation (remembering that the cross term from the squared gradient will vanish)

$$\left\{  \nabla'^2_\perp + \gamma^2 \left( \nabla'_\parallel - \frac{ 1 }{c^2} \vec v \pdiff{}{t'} \right)^2 - \gamma^2 \left( \pdiff{}{t'} -  \vec v \cdot \nabla'_\parallel \right)^2 \right\} \phi = 0$$

Evaluating the terms in parentheses separately

\begin{eqnarray*}
\left( \nabla'_\parallel - \frac{ 1 }{c^2} \vec v \pdiff{}{t'} \right)^2 &=& \delsqprime_\parallel - \frac{ 1 }{ c^2 } ( \vec v \cdot \nabla'_\parallel ) \pdiff{}{t'} - \frac{ 1 }{ c^2 } ( \cancel{ \nabla'_\parallel \cdot \vec v } ) \pdiff{}{t'} - \frac{1}{c^2} (\vec v \cdot \nabla'_\parallel ) \pdiff{}{t'} + \frac{v^2}{c^4} \pdiffsq{}{t'} \\
&=&  \delsqprime_\parallel - \frac{ 2 }{ c^2 } ( \vec v \cdot \nabla'_\parallel ) \pdiff{}{t'} + \frac{v^2}{c^4} \pdiffsq{}{t'} 
\end{eqnarray*}

where one of the cross terms is expanded according to the product rule, and the quantity $\div \vec v$ vanishes. The squared time derivative is

$$\left( \pdiff{}{t'} -  \vec v \cdot \nabla'_\parallel \right)^2 = \pdiffsq{}{t'} - 2 (\vec v \cdot \nabla'_\parallel) \pdiff{}{t'} + (\vec v \cdot \nabla'_\parallel)^2 $$

The wave equation involves the difference of these terms, with the time operator multiplied by $1/c^2$

\begin{eqnarray*}
\left( \nabla'_\parallel - \frac{ 1 }{c^2} \vec v \pdiff{}{t'} \right)^2 - \frac{1}{c^2} \left( \pdiff{}{t'} -  \vec v \cdot \nabla'_\parallel \right)^2 &=& \delsqprime_\parallel - \frac{1}{c^2}(\vec v \cdot \nabla'_\parallel)^2 + \frac{v^2}{c^4} \pdiffsq{}{t'} - \frac{1}{c^2} \pdiffsq{}{t'} \\
&=&  \left( 1 - \frac{v^2}{c^2} \right) \delsqprime_\parallel - \frac{1}{c^2} \left( 1 - \frac{v^2}{c^2} \right) \pdiffsq{}{t'} \\
&=& \left( 1 - \frac{v^2}{c^2} \right) \left( \delsqprime_\parallel - \frac{1}{c^2} \pdiffsq{}{t'} \right)
\end{eqnarray*}

note that the mixed term (involving both time and space derivatives) vanishes, and since $\nabla'_\parallel$ is parallel to $\vec v$, $(\vec v \cdot \nabla'_\parallel)^2 = v^2 \nabla'^2_\parallel$. Finally if we insert this into the wave equation

$$\left\{  \nabla'^2_\perp + \gamma^2 \left( 1 - \frac{v^2}{c^2} \right) \left( \delsqprime_\parallel - \frac{1}{c^2} \pdiffsq{}{t'} \right) \right\} \phi = 0$$

then if $\gamma$ is the usual Lorentz factor, we have recovered the wave equation in the transformed coordinates. Crucially, these waves also travel at speed $c$, even though we are now in the \emph{moving} medium.

\section{Notes on Lorentz Transform}

\subsection{Simultaneity}

Lorentz thought of $t'$, the local time, as simply a change of variable which yielded a wave equation. He did not interpret it physically.

Poincare (1900) noticed that the local time is exactly the type of time that could be used to define simultaneity in two different moving frames.

\subsection{Fizeau}

\section{Four Vectors}

Consider space and time on the same footing. Mathematically, this means that space and time should `belong' to the same object.

In the three dimensions of space, we have 3-vectors $\vec x$. In space-time, we define the \emph{covariant 4-vector}

$$x_\mu \equiv \left[ ct, \vec x \right] = (x_0,x_1,x_2,x_3)$$

For the purposes of this course, we will use the convention of writing $x_\mu$ with a downstairs index. Most particle physics literature reverses this convention. The upstairs/downstairs distinction is arbitrary, but must be applied consistently.

We then define the \emph{contravariant 4-vector}

$$x^\mu \equiv (ct, - \vec x)$$

We define the square of a 4-vector

$$x^2 = x_\mu x^\mu \equiv c^2 t^2 - \vec x^2$$

using the Einstein summation convention. This quantity is invariant under Lorentz transformations. To prove this, we substitute the transformed 3-vector of spatial coordinates, and the transformed time coordinate, into the scalar product

\begin{eqnarray*}
x'^2 &=& c^2 t'^2 - \vec x'^2 \\
&=& c^2 \gamma^2 (t - vx/c^2)^2 -  \gamma^2 (x - vt)^2 - y^2 - z^2 \\
&=& c^2 \gamma^2 (1 - v^2 / c^2) t^2 -  \gamma^2 (1 - v^2 / c^2) x^2 - y^2 - z^2 \\
&=& c^2 t^2 - \vec x^2 = x^2
\end{eqnarray*}

where $\gamma^2 (1 - v^2 / c^2) = 1$.

We define the \emph{metric tensor} of \emph{Minkowski} space-time

$$\eta_{\mu \nu} =
\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{array}\right)
$$

similarly

$$\eta^{\mu \nu} =
\left(\begin{array}{cccc}
-1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -1 & 0 \\
0 & 0 & 0 & -1
\end{array}\right)
$$

It is easy to see that $\eta^{\mu \nu}$ is inverse to $\eta_{\mu \nu}$

$$\eta^{\mu \alpha} \eta_{\alpha \nu} = \delta^\mu \, _\nu =
\left(\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right)
$$

We can use $\eta_{\mu \nu}$ and $\eta^{\mu \nu}$ to raise and lower indices

$$x^\mu = \eta^{\mu \nu} x_\nu, \, \, x_\mu = \eta_{\mu \nu} x^\nu$$

thus we can rewrite

$$x^2 = \eta^{\mu \nu} x_\mu x_\nu = \eta_{\mu \nu} x^\mu x^\nu = \delta^\mu \, _\nu x_\mu x^\nu = x_\mu x^\nu \equiv x \cdot x$$

By our conventions, the 4-vector $x_\mu$ absorbs the minus sign. In other conventions it is the metric tensor $\eta_{\mu \nu}$ which is defined with an overall minus sign. This sign choice is only apparent when components are written explicitly.

Note that in this notation contracted indices (those which are summed over) are \emph{always} in upstairs-downstairs pairs. Free indices must always be consistent in position (upstairs or downstairs) across equals signs.

A Lorentz transformation is \emph{linear}, and so can be written as

$$x_\mu \rightarrow x'_\mu = \Lambda_\mu \, ^\nu x_\nu$$

where

$$\Lambda_\mu \, ^\nu =
\left(\begin{array}{cccc}
\gamma & - \gamma \beta & 0 & 0 \\
- \gamma \beta & \gamma & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 \\
\end{array}\right)
$$

and $\beta \equiv v/c$, the \emph{boost factor}.

So the contravariant 4-vector $x^\mu$ transforms as

$$x^\mu \rightarrow x'^\mu = \Lambda^\mu \, _\nu x^\nu$$

The product of covariant and contravariant vectors (the dot product) is invariant

Now if we transform the product of these, the dot product must be invariant

\begin{eqnarray*}
x^2 \rightarrow x'^2 &=& \left( \Lambda^\mu\, _\alpha x^\alpha \right) \left( \Lambda_\mu \, ^\beta x_\beta \right) \\
&=& \Lambda^\mu\, _\alpha \Lambda_\mu \, ^\beta x^\alpha x_\beta = x^2
\end{eqnarray*}

so we can identify the product of Lorentz transformations with the Kronecker delta

$$\Lambda^\mu \, _\alpha \Lambda_\mu \, ^\beta = \delta_\alpha \, ^\beta$$

or, as matrices

$$\Lambda \Lambda^T = I$$

So we see that a Lorentz transform is just an orthogonal rotation, but in space of \emph{negative metric}. To make this explicit, consider the \emph{rapidity} $\eta$, defined by

$$\beta = \tanh \eta < 1$$

$$\cosh \eta = \gamma, \, \, \sinh \eta = \beta \gamma$$

since $ \cosh^2 \eta - \sin^2 \eta = \gamma^2 ( 1 - \beta )^2 = 1$. Therefore we can write

$$\Lambda_\mu^\nu = 
\left(\begin{array}{cccc}
\cosh \eta & - \sinh \eta & 0 & 0 \\
- \sinh \eta & \cosh \eta & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right)
$$

Compare usual spatial rotations in cartesian three-dimensional space

$$x_i \rightarrow R_{ij} x_j$$

$$R R^T = 1$$

with $x^2 = x_i x_i$ invariant. Note that in cartesian space $\eta_{ij} = \delta_{ij} = \eta^{ij}$, so $x_i = x^i$.

$$R_i^j =
\left(\begin{array}{cccc}
\cos \theta & \sin \theta & 0 & 0 \\
- \sin \theta & \cos \theta & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{array}\right)
$$

In fact, spatial rotations are special cases of Lorentz transformations

$$\Lambda_\mu^\nu =
\left(\begin{array}{cc}
1 & 0 \\
0 & R_{ij}
\end{array}\right)
$$

So the Lorentz transform consists of rotations plus boosts. Just as the group of 3-dimensional rotations form a group, the Lorentz transformations form a group.

Besides the position vector $x_\mu$, we can have \emph{vectors} $V_\mu$, \emph{tensors} $T_{\mu \nu}$ etc. Under a Lorentz transform

$$V_\mu \rightarrow \Lambda_\mu^\nu V_\nu$$
$$T_{\mu \nu} \rightarrow \Lambda_\mu^\alpha \Lambda_\nu^\beta T_{\alpha \beta} = \Lambda_{\mu \alpha} \Lambda_{\nu \beta} T^{\alpha \beta}$$

and of course \emph{scalars}, which are invariant under Lorentz transformations.

We can do \emph{vector calculus} with 4-vectors; consider $x'_\mu (x_\mu)$

$$\d x'_\mu = \pdiff{ x'_\mu }{ x_\nu } \d x_\nu = \Lambda_\mu^\nu \d x_\nu$$

so

$$\Lambda_\mu^\nu = \pdiff{x'_\mu}{x_\nu}$$

note that downstairs indices in denominators are the same as upstairs indices in numerators.

By the chain rule, the partial derivative transforms according to

$$\partial{}{x_\mu} = \partial{x'_\nu}{x_\mu} \partial{}{x'_v}$$

now for notational convenience, we define $\partial^\mu \equiv \pdiff{}{x_\mu}$, and so

$$\partial^\mu = \Lambda_\nu^\mu \partial^{\nu'}$$

(notice the prime mark on the $\nu$). So

$$\Lambda^\alpha_\mu \partial^\mu = \Lambda^\alpha_\nu \Lambda_\nu^\mu \partial^{\eta'} = \delta^\alpha_\nu \partial^{\nu'} = \partial^{\alpha'}$$

so

$$\partial^{\mu'} = \lambda^\mu \, _\nu \partial^\nu$$

i.e. $\partial^\mu$ is a contravariant vector (operator). Likewise

$$\partial_\mu \equiv \pdiff{}{x^\mu}$$

is a contravariant vector (operator), and

$$\partial^2 = \partial_\mu \partial^\mu = \partial^\mu \partial_\mu = \eta_{\mu \nu} \partial^\mu \partial^\nu$$

is a scalar (invariant) operator, the Laplacian.

In our chosen notation

$$\partial^\mu = \left( \frac{1}{c} \pdiff{}{t}, \grad \right)$$
$$\partial_\mu = \left( \frac{1}{c} \pdiff{}{t}, - \grad \right)$$

now if we calculate $\partial^2 = \partial_\mu \partial^\mu$ (which is a scalar, and so must be invariant), we obtain

$$\partial^2 = \partial_\mu \partial^\mu = \frac{1}{c^2} \pdiffsq{}{t} - \delsq$$

i.e. a wave operator, which is sometimes denoted by a ``box'', the \emph{d'Alembertian} operator. So the wave equation can be written very compactly as

$$\partial^2 \phi = 0$$

and is unchanged under a Lorentz transformation, i.e.

$$\partial^2 \phi(x) = 0 \rightarrow \partial'^2 \phi(x') = 0$$

which it must be, as this was our starting point (i.e. we obtained the Lorentz transformation by considering the invariance of the wave equation).

For rotations, there are two invariant tensors, $\delta_{ij}$ and $\epsilon_{ijk}$ (i.e. the \emph{isotropic} tensors).

For Lorentz transformations, there are also two invariant tensors, $\delta_\mu \, ^\nu$ and $\epsilon_{\mu \nu \alpha \beta}$, where $\epsilon_{\mu \nu \alpha \beta}$ is a totally antisymmetric tensor (so $\epsilon^{0123} = \epsilon_{1230} = -1$).

With these invariant tensors and the differential $\d x_\mu$, we can construct \emph{two} invariant differentials:

\begin{itemize}

\item The \emph{interval}

$$\d s^2 = \d x^2 = \eta^{\mu \nu} \d x_\mu \d x_\nu$$

c.f. the interval in 3 dimensions $\d \vec x^2 = \delta_{ij} \d x_i \d x_j$, and

$$\d^4 x = - \frac{1}{4!} \epsilon^{\mu \nu \alpha \beta} \d x_\mu \d x_\nu \d x_\alpha \d x_\beta = c \d t \d x \d y \d z$$

c.f. the volume element in 3 dimensions $\d^3 x = \frac{1}{3!} \epsilon_{ijk} \d x_i \d x_j \d x_k = \d x \d y \d z$. Note that the 3-dimensional volume element is \emph{not} invariant

\item The \emph{volume element}

$$\d^3 \vec x \rightarrow \d^3 \vec x' = \frac{1}{\gamma} \d^3 \vec x$$

i.e. the 3-volume shrinks under a Lorentz transformation, which is a manifestation of Lorentz contraction.

We have the 4-dimensional analogue of Gauss' Theorem

$$\int_V \partial^\mu V_\mu \d^4 x = \oint_S V_\mu \d \Sigma^\mu$$

where $\d \Sigma^\mu$ is a 3-dimensional surface element. The analogue of Stokes' Theorem is much more complicated, as the generalisation of the cross-product to higher dimensions requires the use of differential forms.

\end{itemize}

\section{Geometry in Minkowski Space}

Let the point $O = (0,0,0,0)$ be the origin of Minkowski space. The interval $\d s^2 = 0$ (with respect to $O$) defines a surface in 4 dimensions, the \emph{light cone}.

For any displacement vector $V_\mu$ (with respect to $O$)

\begin{itemize}

\item if $\d s^2 = V_\mu V^\mu < 0$, the vector is called \emph{space-like}.

\item if $\d s^2 = V_\mu V^\mu = 0$, the vector is called \emph{null} (note that null 3-vectors necessarily have all null components, which is not the case with 4-vectors).

\item if $\d s^2 = V_\mu V^\mu > 0$ the vector is called \emph{time-like}.

\end{itemize}

The `type' of a vector is determined by $\d s^2$, which is an invariant, hence under any Lorentz transformation the `type' of a vector is preserved:

\begin{itemize}

\item time-like $\rightarrow$ time-like

\item space-like $\rightarrow$ space-like

\item time-like $\not \leftrightarrow$ space-like

\end{itemize}

i.e. there is \emph{no} Lorentz transformation which takes space-like vectors into time-like vectors, or vice-versa.

Observers can only exchange light rays (i.e. communicate) if they are separated by a time-like interval. The $(+,-,-,-)$ convention for the metric ensures that $\d s = \sqrt{ \d s^2 }$ is real for time-like intervals (i.e. intervals which can actually be traversed), but becomes imaginary for space-like intervals.

\subsection{Orthogonality}

Angles, and orthogonality, can be defined in the usual way (i.e. with respect to the inner product). Two 4-vectors $V$ and $W$ are called orthogonal if

$$V_\mu W^\mu = 0$$

(i) If $V_\mu V^\mu = 0$ (i.e. a null vector), $V_\mu$ is called \emph{self-orthogonal}. Note that for 3-vectors $v_i^2 = 0 \Rightarrow v_i = 0$, but this is not true for 4-vectors.

(ii) If $V_\mu$ is time-like, and $V_\mu W^\mu = 0$, then $W_\mu$ is space-like.

(ii) If $V_\mu$ is null, and $V_\mu W^\mu = 0$, then $W_\mu$ is either null or space-like (i.e. not timelike).

In order to prove the above statements, realise that any null vector is proportional to $(1,0,0,1)$ (by choosing the appropriate frame), whereas any time-like vector is proportional to $(1,0,0,0)$, and space-like vector is proportional to $(0,0,0,1)$.

\subsection{Example: The Doppler Effect}

Let $f$ be a Lorentz scalar, which satisfies the wave equation

$$\partial^2 f = 0$$

$f$ can always be decomposed into plane waves $f = e^{i \phi}$. Since $f$ is a Lorentz scalar, then so must be $\phi$. Now consider the action of $\partial^2$ on $f$

\begin{eqnarray*}
\partial^2 e^{i \phi} &=& \partial^\mu \left[ \partial_\mu e^{i \phi} \right] \\
&=& \partial^\mu \left[ i \partial_\mu (  \phi ) e^{i \phi} \right] \\
&=& \left[ i \partial^2 ( \phi ) - \partial^\mu ( \phi ) \partial_\mu ( \phi ) \right] e^{i \phi} = 0
\end{eqnarray*}

now for the original wave equation to be satisfied, the terms in brackets must vanish, hence

$$ i \partial^2 ( \phi ) = \partial^\mu (\phi) \partial_\mu (\phi) $$

i.e. we have a differential equation for $\phi$. If we try a plane wave ansatz

$$\phi = k^\mu x_\mu$$

which is linear in $x_\mu$, then $\partial^2 \phi = 0$ trivially, leaving

$$\partial_\mu ( k_\mu x^\mu ) \partial^\mu ( k^\mu x_\mu ) = k_\mu k^\mu = 0$$

i.e. the differential equation is satisfied is $k_\mu$ is null. Now by construction $\phi = k_\mu x^\mu$ is a Lorentz scalar, and since $x^\mu$ is a 4-vector, then by the quotient theorem $k_\mu$ is a null 4-vector.

$$k_\mu = \left( \frac{\omega}{c} , \vec k \right)$$

$\omega = c k$ (where $k = | \vec k |$). $\omega$ is the frequency (with dimensions of $1/t$, $k^\mu x_\mu = \omega t - \vec k \cdot \vec x$), $| \vec k |$ is the wave number, $\hat k$ gives the direction of propagation of the wave.

Since $\omega$ is not itself a scalar, but the component $k_0$ of a 4-vector $k_\mu$, then under a Lorentz transform

$$ \frac{\omega}{c} \rightarrow \frac{\omega'}{c} = \gamma \left( \frac{\omega}{c} + \vec k \cdot \vec v / c \right) $$

i.e. $\omega' = \gamma ( \omega + \vec k \cdot \vec v )$, i.e. the frequency is shifted by a change of reference frame; this is the relativistic Doppler effect.

\begin{itemize}

\item if $\vec k \cdot \vec v < 0 \Rightarrow \omega' < \omega$

i.e. a red shift when the source and observer are moving away from one another (the wave-vector and frame velocity are anti-aligned).

\item If $\vec k \cdot \vec v > 0 \Rightarrow \omega' > \omega$

i.e. a blue shift when the source and observer and moving towards one another (the wave-vector and frame velocity are aligned).

\end{itemize}

\section{Relativistic Dynamics}

In Newtonian dynamics, the position of a particle is described by $$x_i(t),$$ its momentum is given by $$p_i = m \diff{x_i}{t},$$ and the force on the particle is $$F_i = \diff{p_i}{t}.$$

These are Newton's equations, and contain all of Newtonian dynamics. Newton's equations are invariant under

\begin{itemize}

\item Rotations $x_i \rightarrow R_{ij} x_j$, $p_i \rightarrow R_{ij} p_j$, $F_i \rightarrow R_{ij} F_j$, $t \rightarrow t$

\item Galilean transformations $x_i \rightarrow x_i - v_i t$

\end{itemize}

but they are \emph{not} invariant under Lorentz transformations $x_\mu \rightarrow \Lambda_\mu \, ^\nu x_\nu$.

The difficulty with Newtonian dynamics is that, under a Lorentz transform, both $x$ and $t$ transform together. We can see that the quantities which result from taking derivatives with respect to $t$, such as

$$v_i = \diff{x_i}{t}$$

are clearly not components of 4-vectors, hence Newtonian dynamics cannot be correct in the relativistic regime.

\subsection{Proper Time}

The time $t$ is not Lorentz invariant. In order to define 4-vectors analogous to Newtonian momentum and acceleration, we must differentiate 4-position $x_\mu$ with respect to some invariant parameter, which is linearly related to ordinary time $t$. 

To find a suitable invariant parameter, start with the interval $$\d s^2 = \eta_{\mu \nu} \d x^\mu \d x^\nu = c^2 \d t^2 - \d \x^2$$ which \emph{is} Lorentz invariant (by construction of the Lorentz transform). We can re-write this as $$\d s^2 = c^2 \left[ 1 - \frac{1}{c^2} \left( \diff{\vec x}{t} \right)^2 \right] \d t^2 = c^2 \left( 1 - \frac{\vec u^2}{c^2} \right) \d t^2 = \frac{c^2}{\gamma^2} \d t^2,$$ where we have defined $$\vec u = \diff{ \x}{t}$$

i.e. $\vec u$ is the velocity of the moving frame. Now $\d s^2$ and $c^2$ are invariant, so by the quotient theorem so must be $\d t^2 / \gamma^2$, thus we define the Lorentz invariant \emph{proper time}

$$\d \tau = \frac{1}{\gamma} \d t$$

so along a particle path ( i.e. $x_i$ is such that $\diff{x_i}{t} = u_i$ ), $\d s^2 > 0$ (i.e. $\d x_\mu$ is time-like). Then

$$ \d \tau = \frac{1}{\gamma} \d t$$

called the \emph{proper time} is Lorentz invariant.

The \emph{``world line''} of the particle (the analogue of a Newtonian trajectory) is $x_\mu ( \tau )$, a 4-vector under Lorentz transform.

The \emph{4-velocity} $$\diff{x_\mu}{\tau} \equiv u_\mu$$ is thus also a 4-vector. In components $$\diff{x_\mu}{\tau} = \gamma \diff{}{t} ( ct, \x ) = \gamma(c,\vec u)$$ so $$u_\mu u^\mu = \gamma^2 \left( c^2 - \vec u^2 \right) = \gamma^2 \left( 1 - \frac{\vec u^2}{c^2} \right) c^2 = c^2$$

i.e. $u_\mu$ is a time-like 4-vector with fixed length c (N.B. when dealing with 4-vectors, remember that $u_\mu u^\mu = u^2 \neq \vec u^2$).

Similarly we can define \emph{4-acceleration} $$a_\mu \equiv \diff{u_\mu}{\tau} = \diffsq{x_\mu}{\tau}$$ so using the above result that $u_\mu u^\mu = c^2$ \begin{eqnarray*}
\diff{}{\tau} ( u_\mu u^\mu ) &=& \diff{}{\tau} c^2 \\
\Rightarrow \diff{u_\mu}{\tau} u^\mu +  u_\mu \diff{u^\mu}{\tau} &=& 0 \\
\Rightarrow a_\mu u^\mu +  u_\mu a^\mu &=& 0 \\
\Rightarrow 2 a_\mu u^\mu = 0
\end{eqnarray*} $u^\mu a_\mu = 0$, i.e. $u_\mu$ and $a_\mu$ are orthogonal. Since $u_\mu$ is time-like, the acceleration 4-vector $a_\mu$ is space-like.

\subsection{Energy and Momentum}

Properties that are `carried with' a particle must be frame-invariant; they are \emph{intrinsic}. Assuming that each particle has a mass $m$, invariant under Lorentz transformations, then we can define \emph{4-momentum} $$p_\mu = m u_\mu = m \diff{x_\mu}{\tau}$$ which is a time-like vector, and again using the result that $u_\mu u^\mu = c^2$ we have $$p^2 = p_\mu p^\mu = m^2 u_\mu u^\mu = m^2 c^2.$$

To interpret the components of $p_\mu$, consider the low-velocity correspondence limit $u \ll c$ in which we expect to recover Newtonian dynamics. Using the binomial expansion on the zeroth component of 4-momentum $$p_0 = \gamma mc = \frac{mc}{\sqrt{ 1 - u^2 / c^2 }} = mc \left[ 1 + \frac{1}{2} \frac{u^2}{c^2} + \order{ \frac{u^4}{c^4} } \right]$$ so if we multiply by $c$ $$p_0 c \approx \underbrace{mc^2}_\textrm{rest energy} + \underbrace{ \frac{1}{2} mu^2}_\textrm{kinetic energy} + \cdots$$ hence we can identify $p_0 c \equiv E$ with the \emph{energy} of the particle. Similarly for the latter components $\mu = i \in [1,2,3]$ $$p_i = \gamma m u_i \approx m u_i + \cdots$$ so we identify $p_i \equiv \vec p$ with the \emph{momentum} of the particle.

We have seen that Newtonian energy and momentum are just low-velocity approximations to the ``exact'' quantities which arise in Special Relativity. Making the above identifications, we can re-write 4-momentum as $$p_\mu = \left( \frac{E}{c}, \vec p \right).$$ In Newtonian dynamics energy and momentum are related by $$E = \frac{\vec p^2}{2m},$$ the relativistic relationship can be obtained by considering the squared 4-momentum $$p^2 = \frac{E^2}{c^2} - \vec p^2$$ but recalling our result that $p^2 = m^2 c^2$, we obtain Einstein's famous mass-energy relation $$\frac{E^2}{c^2} - \vec p^2 = m^2 c^2 \Rightarrow E = \sqrt { \vec p^2 c^2 + m^2 c^4 }.$$

\subsection{Energy Momentum Conservation}

$p^\mu$ is conserved, e.g.

$$p_1^\mu + p_2^\mu = p_3^\mu + p_4^\mu$$

Newton's Law then becomes

$$\diff{p_\mu}{\tau} = F_\mu$$

where $F_\mu$ is a space-like 4-vector, and since $\diff{p_\mu}{\tau}$ is proportional to $a_\mu$, then $F_\mu$ must be orthogonal to $u_\mu$

$$F_\mu u^\mu = 0$$

Hence note that only 3 component of $F_\mu$ are independent: these are

$$F_i = \diff{}{\tau}p_i = \gamma \diff{}{\tau} ( \gamma m u_i )$$

It is easy to see that, since $F_\mu u^\mu = 0$, and $u^\mu = \gamma( c, - \vec u)$, then

$$F_0 = \frac{1}{c} \vec F \cdot \vec u$$

i.e. the power expended is

$$\diff{E}{\tau} = \vec F \cdot \vec u$$

\subsection{Massless Particles}

Relativistic dynamics allows us to consider massless particles (i.e. $m \rightarrow 0$), which would not make sense in Newtonian mechanics. If we take the mass-energy relationship, and set $m=0$, we obtain $$E^2 = \vec p^2 c^2.$$ Observe that, substituting this into the momentum 4-vector $$p_\mu = \left( \frac{E}{c}, \vec p \right) = \left( \frac{| \vec p | c}{c} , \vec p \right) = \left( | \vec p | , \vec p \right) \Rightarrow p^2 = 0$$ i.e. for massless particles $p_\mu$ is null.

This is only possible if $u^2 \rightarrow c^2$. To show this, consider that for any particle $$E = \gamma mc^2 = \frac{mc^2}{\sqrt{ 1 - u^2 / c^2}}$$ now we want to take the limit $m \rightarrow 0$, but keeping $E$ fixed and non-zero. To take this limit, we must first rearrange $$E^2 \left( 1 - \frac{u^2}{c^2} \right) = m^2 c^4$$ $$\Rightarrow 1 - \frac{u^2}{c^2} = \frac{m^2 c^4}{E^2}$$ $$\Rightarrow u^2 = c^2 \left[ 1 - \left( \frac{mc^2}{E} \right)^2 \right]$$ then if we take $m \rightarrow 0$, the term in parentheses vanishes and we are left with $$u = c$$ i.e. massless particles always travel at the speed of light.

\section{Relativistic Electrodynamics}

To build up relativistic dynamics, we have had to amend Newton's laws, and we have seen along the way that we maintain correspondence in Newtonian limits. To write relativistic electrodynamics, we will essentially be transcribing Maxwell's equations into relativistic notation.

\subsection{Electric Charge}

Assume that a particle has \emph{electric charge}, $q$, which is Lorentz invariant. Just like the mass, this is an \emph{intrinsic} quality. This must be true physically, as otherwise when electrons move, their charge would change. Consider a wire with electrons $+,-$, when a current is applied the negative charges move, and if charge were not a Lorentz scalar, then the wire would acquire an overall charge, which does not happen.

Then the current due to a \emph{single particle} is

$$j_\mu = q u_\mu = q \gamma (c,\vec u)$$

which is a time-like 4-vector

$$j^2 = q^2 c^2$$

When $u \ll c$,  $\gamma \approx 1$, then we can identify

$$j_\mu = \left( \underbrace{q c}_\textrm{charge} , \underbrace{q \vec u}_\textrm{current} \right)$$

Similarly we define \emph{current density}

$$J_\mu ( x ) = \left( \underbrace{ \rho( \x, t) }_\textrm{charge density} c, \J ( \x ,t ) \right) $$

such that 

$$J_\mu = \rho_0 u_\mu = \rho_0  \gamma(c,\vec u)$$

i.e. $\rho = \gamma \rho_0$, $\J = \gamma \rho_0 \vec u$, where $\rho_0$ is the (Lorentz invariant) \emph{proper charge density}: in the rest frame of a moving charge

$$q = \int_V \rho \d^3 \x$$

then in any frame

$$q = \int_V \rho \d^2 \x = \int_V \rho_0 \gamma \d^3 \x$$

this works because $\gamma d^3 \x$ is invariant. Remember that $\d^4 x = c \d t \d^3 \x$, and $\d t = \gamma \d \tau$, and the proper time is an invariant. Since $\d^4 x$ and $\d \tau$ are invariant, if we substitute for $\d t$ in $\d^4 x$, we obtain $\d^4 x = c \gamma \d \tau \d^3 \x$, and so $\gamma \d^3 \x$ must be invariant.

Since $J_\mu$ is a 4-vector under Lorentz transformation, $J_\mu \rightarrow \Lambda_\mu \, ^\nu J_\nu$, so

$$\rho \rightarrow \rho' = \gamma \left( \rho - J_\parallel \frac{v}{c^2} \right)$$
analogously
$$t \rightarrow \gamma \left( t - x_\parallel \frac{v}{c^2} \right)$$

Similarly, for the parallel component of $J$

$$J_\parallel \rightarrow J'_\parallel = \gamma \left( J_\parallel - \rho v \right)$$
analogously
$$x_\parallel \rightarrow x'_\parallel \gamma \left( x_\parallel - vt \right)$$

and perpendicular components are unchanged

$$J_\perp \rightarrow J'_\perp = J_\perp$$
analogously
$$x_\perp \rightarrow x_\perp$$

so if $v \ll c$

$$\rho' \approx \rho$$
$$J'_\parallel \approx J_\parallel - \rho v$$
$$J'_\perp \approx J_\perp$$

just as expected.

But for $v < \sim c$, we get relativistic corrections, just as for space-time.

\subsection{Example (Feynman)}

This example will illustrate the relationship between magnetic and electric fields, and is due to Feynman. Consider a wire, with charge densities $\rho^+$ and $\rho^-$, with $\rho^\pm = \pm \rho_0$, so there is no net charge.

In frame $S$, $\rho^-$ moves with velocity $\vec v$, and $\rho^+$ is stationary.

The current in the wire is $$\J^- = \rho^- \vec v = - \rho_0 \vec v,$$ and hence there is a solenoidal magnetic field around the wire of $$\B = - \frac{v \rho A}{2 \pi r c} \hat{\vec \phi},$$ which we can find using Ampere.

If we have a test charge $q$ moving with velocity $\vec v$, the force experienced by the test charge in the frame $S$ is given by the Lorentz force law $$\vec F = q \frac{1}{c} \vec v \times \B = \frac{q \rho_0 A}{2 \pi r} \left( \frac{v}{c} \right)^2 \hat \r.$$

Now if we perform a Galilean transformation to a different frame $S'$, chosen such that $\rho^-$ is stationary; $\rho^+$ moves with velocity $-\vec v$, and the test charge $q$ is stationary. In frame $S'$, $q$ is stationary, and so the Lorentz force $\vec F'$ on $q$ will vanish. However, the force vector is orthogonal to the velocity of the moving frame $S'$, so it should be unchanged. It seems as though real, physical forces can be made to vanish just by a choice of frame; this is a paradox.

The caveat is that we used a Galilean transform on $\vec v$ and $\rho^\pm$, where a Lorenz transformation is called for. If we perform the correct transformation

$$\rho'^+ = \gamma \rho^+ = \gamma \rho_0$$

$$\rho'^- = \gamma \left( \rho^- - \J^- \frac{v}{c^2} \right) = \gamma \left( 1 - \frac{v^2}{c^2} \right) \rho^- = - \frac{1}{\gamma} \rho_0$$

So the resulting charge density in the wire is

$$\rho'^+ + \rho'^- = \left( \gamma - \frac{1}{\gamma} \right) \rho_0 = \gamma \left( 1 - \frac{v^2}{c^2} \right) \rho_0 = \gamma \left( \frac{v}{c} \right)^2 \rho_0$$

and the electric field is

$$\E' = \gamma \left( \frac{v}{c} \right)^2 \frac{\rho_0 A}{2 \pi r} \hat \r$$

and so the force on the (now stationary) charge $q$ is

$$\vec F' = \gamma \frac{q \rho_0 A}{2 \pi r} \left( \frac{v}{c} \right)^2 \hat \r$$

c.f. $\vec F$, we acquire a factor of $\gamma$, i.e. $\vec F' = \gamma \vec F$. Note that this is \emph{not} the proper force (i.e. the last 3 components of 4-force), hence it acquires a factor of $\gamma$.

So, the magnetic force on $q$ may be thought of as purely a relativistic effect. Note that is is $\order{ \frac{v^2}{c^2} }$ (since $\gamma$ only differs from unity by terms $\order{ \frac{v^2}{c^2} }$). The magnetic force is still observable however, even when $v \ll c$, since the charges are huge.

\subsection{Charge Consevation}

Charge conservation is typically stated as

$$\pdiff{\rho}{t} + \div \J = 0$$

i.e. since

$$\partial^\mu = \left( \frac{1}{c} \pdiff{}{t}, \grad \right), \, J_\mu = \left( \rho c, \J \right)$$

then

$$\partial^\mu J_\mu = \pdiff{\rho}{t} - \div \J$$

is simply a restatement of charge conservation. Now since physically, charge must be conserved in all frames, the above must be an invariant. Now by the quotient theorem since $\partial^\mu$ is a 4-vector, then so must be $J_\mu$

Now we apply the 4-dimensional version of the divergence theorem: for any closed space-time volume $V_4$

$$0 = \int_{V_4} \partial^\mu J_\mu \d^4 x = \oint_{S_3} \d \Sigma^\mu J_\mu$$

If we take $V_4$ to be a 4-dimensional cylinder, oriented as follows

then on the ends of the cylinder $$\d \Sigma_\mu = ( \d^3 \x , 0 )$$, and on the sides of the cylinder $$\d \Sigma_\mu = ( 0, - \d \S c \d t )$$, so we can perform this integral

$$c \int_{V_3} \d^3 \x \left( \underbrace{ \rho(\x,t) }_\textrm{top} - \underbrace{ \rho(\x,0) }_\textrm{bottom} \right) + c \int_0^t \d t' \oint_{S_2} \d \S \cdot \J(\x, t')$$

Differentiate this w.r.t. $t$ to obtain

$$\diff{}{t} \int_{V_3} \d^3 \x \rho(\x,t) = - \oint_{S_2} \d \S \cdot \J(\x,t) = 0$$

\subsection{Electromagnetic Potential}

We showed previously that in Lorentz gauge

$$\left( \delsq - \frac{1}{c^2} \pdiffsq{}{t} \right) \phi = - \rho$$

$$\left( \delsq - \frac{1}{c^2} \pdiffsq{}{t} \right) \A = - \frac{1}{c} \J$$

But

a) $\delsq - \frac{1}{c^2} \pdiffsq{}{t} = - \partial^2 \equiv - \partial_\mu \partial^\mu$ is a Lorentz scalar (operator)

b) $J_\mu = ( \rho c, \J )$ is a 4-vector

It follows that $A^\mu (x) = \left( \phi(\x,t), \A(\x,t) \right)$ is a vector, and

$$\partial^2 A_\mu = \frac{1}{c} J_\mu$$

i.e. the content of Maxwell's equations is that 

To get here, we needed to use the Lorenz gauge condition, i.e.

$$\frac{1}{c} \partial{}{t} \phi + \div \A = 0$$

i.e. $\partial^\mu A_\mu = 0$, the 4-divergence of $A_\mu$ is zero. This gauge condition is \emph{also} Lorentz invariant; this is why it was special. The gauge transforms

$$\phi \rightarrow \phi - \frac{1}{c} \pdiff{}{t} \chi$$

$$\A \rightarrow \A + \grad \chi$$

may also be written

$$A_\mu \rightarrow A_\mu + \partial_\mu \chi$$

so the gauge function $\chi(x)$ is \emph{also} a Lorenz invariant scalar.

\subsection{Electromagnetic Fields}

In terms of the potentials, the fields $\E$ and $\B$ are

$$\E = - \frac{1}{c} \pdiff{}{t} \A - \grad \phi, \; \B = \curl \A$$

Consider the following object

$$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

which is a sort of cross-product in 4 dimensions ``$\mathrm{curl}(A_\mu)$'', where $A_\mu = (\phi,\A)$. Now

\begin{itemize}
\item if $\mu = i, \; \nu = 0$ $$F_{i0} = \partial_i A_0 - \partial_0 A_i = - \nabla_i \phi - \frac{1}{c} \pdiff{}{t} A_i = E_i$$
\item if $\mu = i, \; \nu = j$ $$F_{ij} = \partial_i A_j - \partial_j A_i = -\epsilon_{ijk} B_k$$ because $$\epsilon_{ijk} B_k = \epsilon_{ijk} \epsilon_{klm} \nabla_l A_m = (\delta_{il} \delta_{jm} - \delta_{lm} \delta_{il} ) \nabla_l A_m = \nabla_i A_j - \nabla_j A_i = - \partial_i A_j + \partial_j A_i$$
\end{itemize}

Note that $F_{\mu \nu} = - F_{\nu \mu}$, so $F_{00} = Fii = 0$. In matrix notation

$$F_{\mu \nu} = \left(\begin{array}{cccc}0 & -E_1 & -E_2 & -E_3 \\E_1 & 0 & -B_3 & B_2 \\E_2 & B_3 & 0 & -B_1 \\E_3 & -B_2 & B_1 & 0\end{array}\right)$$

so the electric and magnetic fields combine to form an antisymmetric rank 2 tensor. Note that we also have the contra-tensor

$$F^{\mu \nu} = \left(\begin{array}{cccc}0 & E_1 & E_2 & E_3 \\ -E_1 & 0 & -B_3 & B_2 \\ -E_2 & B_3 & 0 & -B_1 \\ -E_3 & -B_2 & B_1 & 0\end{array}\right)$$

Note that $F_{\mu \nu}$ is gauge invariant, as it should be, since it is just an alternative way of writing the electric and magnetic fields. If $$A_\mu \rightarrow A_\mu + \partial_\mu \chi,$$ then $$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\mu A_\nu \rightarrow F_{\mu \nu} + \partial_\mu \partial_\nu \chi - \partial_\nu \partial_\mu \chi = F_{\mu \nu}$$

We can now recover the Maxwell equations

\begin{eqnarray*}
\partial^\mu F_{\mu \nu} &=& \partial^\mu (\partial_\mu A_\nu - \partial_\nu A_\mu) \\
&=& \partial^2 A_\nu - \partial_\nu (\partial^\mu A_\mu)
\end{eqnarray*}

so $$\partial^\mu F_{\mu \nu} = \frac{1}{c} J_\nu$$ since in Lorentz gauge $\partial^2 A_\mu = \frac{1}{c} J_\mu$ and $\partial^\mu A_\mu = 0$, or in components

if $\nu = 0$ $$\div \E = \rho$$

if $\nu = i$
\begin{eqnarray*}
\frac{1}{c} J_i &=& \partial^0 F_{0i} + \partial^j F_{ji} \\
&=& - \frac{1}{c} \pdiff{}{t} E_i - \underbrace{ \nabla_j \epsilon_{ijk} B_k }_{-\curl \B}
\end{eqnarray*}
i.e. $$\curl \B = \frac{1}{c} \J + \frac{1}{c} \pdiff{}{t} \E$$ as required.
Note that charge conservation is automatically satisfied $$\frac{1}{c} \partial^\nu J_\nu = \partial^\nu \partial^\mu F_{\mu \nu} = 0$$

For the other two (kinematic) equations, we define a \emph{dual} field strength

$$F^\star_{\mu \nu} = \frac{1}{2} \epsilon_{\mu \nu \alpha \beta} F^{\alpha \beta}$$

where $\epsilon_{\mu \nu \alpha \beta}$ is a totally antisymmetric tensor with 4 indices. To find the components of $F^\star_{\mu \nu}$ we must perform manipulations on the antisymmetric tensor. First note that $\epsilon_{ijk0} = \epsilon_{ijk}$, but care must be taken to perform permutations individually, as the indices do not simply cycle round, e.g. $\epsilon_{0ijk} \neq \epsilon_{ijk0}$, but rather: $$\epsilon_{ijk0} = - \epsilon_{0jki} = \epsilon_{0ikj} = - \epsilon_{0ijk}$$

then

$$F^\star_{0i} = \frac{1}{2} \epsilon_{0ijk} F^{jk} = - \frac{1}{2} \epsilon_{ijk} \epsilon_{jkl} B_k = \frac{1}{2} ( \delta_{ij} \delta_{jl} - \delta_{il} \delta_{jj})B_l = - B_i$$

while

$$F^\star_{ij} = \epsilon_{ijk0} F^{k0} = \epsilon_{ijk} E_k$$

so

$$F^\star_{\mu \nu} =  \left(\begin{array}{cccc}0 & -B_1 & -B_2 & -B_3 \\B_1 & 0 & E_3 & -E_2 \\B_2 & -E_3 &  & E_1 \\B_3 & E_2 & -E_1 & 0\end{array}\right)$$

$$[ \epsilon_{ijk0} = - \epsilon_{0ijk} = - \epsilon_{ijk} ]$$

thus under duality

$$F_{\mu \nu} \rightarrow F^\star_{\mu \nu}, \; \E \rightarrow \B, \; \B \rightarrow - \E$$

Now

\begin{eqnarray*}
\partial^\mu F^\star_{\mu \nu} &=& \frac{1}{2} \epsilon_{\mu \nu \alpha \beta} \partial^\mu ( \partial^\alpha A^\beta - \partial^\beta A^\alpha) \\
&=& - \epsilon_{\mu \nu \alpha \beta} \partial^\mu \partial^\alpha A^\beta = 0
\end{eqnarray*}

i.e. $$\partial^\mu F^\star_{\mu \nu} = 0$$ so in components

if $\nu = 0$, $$\div \B = 0$$

if $\nu = i$,

\begin{eqnarray*}
0 &=& \partial^0 F^\star_{0i} + \partial^j F^\star_{ji} \\
&=&- \frac{1}{c} \pdiff{}{t} B_i - (\curl \E)_i
\end{eqnarray*}

i.e. $$\curl \E = - \frac{1}{c} \pdiff{}{t} \B$$ as required.

Note that when $J_\mu = 0$ (i.e. in the absence of sources) $$\partial^\mu F_{\mu \nu} = \partial^\mu F^\star_{\mu \nu} = 0$$ i.e. we have complete symmetry under duality $$\E \rightarrow \B, \; \B \rightarrow - \E$$

The source-free Maxwell equations can be written

$$\partial^\mu F^\star_{\mu \nu} = - \frac{1}{2} \epsilon_{\nu \mu \alpha \beta} \partial^\mu F^{\alpha \beta} = 0$$ so colloquially ``$\mathrm{curl}(F) = 0$''

If we fix $\nu$, then $$\partial_\mu F_{\alpha \beta} + \partial_\alpha F_{\beta \mu} + \partial_\beta F_{\mu \alpha} = 0$$ this is called the ``Bianchi identity''. We can check this by substituting $$F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$$

\subsection{Lorentz Transformation of Electric and Magnetic Fields}

Since $F_{\mu \nu}$ is a tensor $$F_{\mu \nu} \rightarrow \Lambda_\mu \, ^\alpha \Lambda_\nu \, ^\beta F_{\alpha \beta}$$ under a Lorentz transform. Consider the usual L.T. in the $x$ direction $$\Lambda_0 \, ^0 = \Lambda_1 \, ^1 = \gamma, \; \Lambda_0 \, ^1 = \Lambda_1 \, ^0 = - \beta \gamma, \; \Lambda_2 \, ^2 = \Lambda_3 \, ^3 = 1$$ and all other components are zero. Then

\begin{eqnarray*}
E_1 &=& F_{10} \Rightarrow \Lambda_1 \, ^0 \Lambda_0 \, ^1 F_{01} + \Lambda_1 \, ^1 \Lambda_0 \, ^0 F_{10} \\
&=& \beta^2 \gamma^2 ( - E_1) + \gamma^2 E_1 = E_1
\end{eqnarray*}
\begin{eqnarray*}
E_2 &=& F_{20} \Rightarrow \Lambda_2 \, ^2 \Lambda_0 \, ^0 F_{20} + \Lambda_2 \, ^2 \Lambda_0 \, ^1 F_{21} \\
&=& \gamma (E_2 - \beta B_3)
\end{eqnarray*}
\begin{eqnarray*}
E_3 &=& F_{30} \Rightarrow \Lambda_3 \, ^3 \Lambda_0 \, ^0 F_{30} + \Lambda_3 \, ^3 \Lambda_0 \, ^1 F_{31} \\
&=& \gamma (E_3 - \beta B_2)
\end{eqnarray*}

i.e. $$\E_\parallel \rightarrow E_\parallel, \; \E_\perp \rightarrow \gamma (\E_\perp + \frac{1}{c} \vec v \times \B),$$ since $$(\vec v \times \B)_2 = - v B_3, \; (\vec v \times \B)_3 = v B_2.$$

Similarly, using the duality transformation $\E \rightarrow \B, \; \B \rightarrow - \E$, we immediately obtain $$\B_\parallel \rightarrow \B_\parallel, \; \B_\perp \rightarrow \gamma( \B_\perp - \frac{1}{c} \vec v \times \E)$$

Note $\vec v \times \B_\perp = \vec v \times \B$, since the entire $\B$ field is perpendicular to $\E_\parallel$ and hence perpendicular to $\vec v$ also.

\subsection{Observations}

\begin{itemize}
\item These are relativistic generalisations of the (nonrelativistic) results $$\E \rightarrow \E + \frac{1}{c} \vec v \times \B = \E', \; \B \rightarrow \B - \frac{1}{c} \vec v \times \E = \B'.$$
\item If in frame $S$, $\B = 0$, then in $S'$ $$\B' = - \frac{1}{c} \vec v \times \E'$$
Similarly if in $S$, $\E = 0$, then in $S'$ $$\E' = \frac{1}{c} \vec v \times \B'$$
i.e. fields generated by a Lorentz transform are always $\perp$ to $\vec v$
\end{itemize}

\subsection{Example}

What are the fields produced by a uniformly moving point charge $q$, moving with a constant velocity $\vec v$ (in $S$), at the origin?

In a frame $S'$ moving with the charge (the rest frame of the charge), we know that the field from the charge is simply the static Coulomb field, and the magnetic field is absent $$\E' (\r') = \frac{q}{4 \pi} \frac{\r'}{r'^3}, \; \B' (\r') = 0.$$

We assume that the particle moves along the $x$-axis, and the origins of the frames coincide at $t=0$ (i.e. they are in standard configuration), so that $$x' = \gamma(x - vt), \; y' = y, \; z' = z.$$ Then at $t=0$ we obtain $$\r' = (\gamma x, y, z), \; r'^2 = \gamma^2 x^2 + y^2 + z^2,$$ so $$E_x = E'_x, \; E_y = \gamma E'_y, \; E_z = \gamma E'_z.$$ Putting all of this together $$\E = \frac{q}{4 \pi} \frac{1}{(\gamma^2 x^2 + y^2 + z^2)^{3/2}} ( \gamma x, \gamma y, \gamma z).$$ So the electric field is still (remarkably) radial. We can simplify the denominator according to $$\gamma^2 x^2 + y^2 + z^2 = \gamma^2 r^2 + (1-\gamma^2)(y^2 + z^2) = \gamma^2 r^2 (1 - \beta^2 \sin^2 \theta),$$ and so $$\E (\r) = \frac{q}{4 \pi} \frac{1}{\gamma^2} \frac{1}{(1-\beta^2 \sin^2 \theta)^{3/2}} \frac{\r}{r^3},$$ so the transformed field looks like a coulomb field, but with the fractor $$ \frac{1}{\gamma^2} \frac{1}{(1-\beta^2 \sin^2 \theta)^{3/2}}.$$

\begin{itemize}
\item If $\theta \approx 0$, $\E$ is \emph{reduced} by a factor $1/\gamma^2$
\item If $\theta \approx \pi / 2$, $\E$ is \emph{increased} by a factor $1 / \gamma^2 (1-\beta^2)^{3/2} = \gamma$
\end{itemize}

The magnetic field is given by $$\vec B = \frac{1}{c} \vec v \times \E = \frac{q}{4 \pi c} \frac{1}{\gamma^2} \frac{1}{(1-\beta^2 \sin^2 \theta)^{3/2}} \frac{\vec v \times \r}{r^3},$$ for $v \ll c$, this is just the Biot-Savart Law.

For ultra-relativistic motion, $v < \approx c$ $\gamma \gg 1$, $\E$ and $\B$ are only non-vanishing in a plane moving with the charge. In this plane $$\E = \frac{q}{4 \pi} \gamma \frac{\r}{r^3}, \; \B \approx \frac{q}{4 \pi c} \gamma \frac{\vec v \times \r}{r^3}.$$

\subsection{Field Invariants}

Clearly $F_{\mu \nu} F^{\mu \nu}$, and $F^\star_{\mu \nu} F^{\mu \nu}$ are invariants under the Lorentz transform. In terms of $\E$ and $\B$ \begin{eqnarray*} F_{\mu \nu} F^{\mu \nu} &=& 2 F_{i0} F^{i0} + F_{ij} F^{ij} \\ &=& -2 \E^2 + \epsilon_{ijk} B_k \epsilon_{ijl} B_l = -2 (E^2 - B^2).\end{eqnarray*} Similarly $$F^\star_{\mu \nu} F^{\mu \nu} = 2 \B \cdot \E + 2 \E \cdot \B = 4 \E \cdot \B.$$ So $$-\frac{1}{4} F_{\mu \nu} F^{\mu \nu} = \frac{1}{2} ( E^2 - B^2), \; \frac{1}{4} F^\star_{\mu \nu} F^{\mu \nu} = \E \cdot \B,$$ so under a Lorentz transform $E^2 - B^2$ and $\E \cdot \B$ are invariants.

Remarks:
\begin{itemize}
\item if $\E \cdot \B = 0$ in some frame, then $\E \cdot \B = 0$ in all frames.
\item if $E^2 = B^2$ in some frame, then $E^2 = B^2$ in all frames.
\item if $\E \neq 0$, $\B = 0$ in some frame, $E^2 - B^2 > 0$ in all frames, so there is no frame in which $\E = 0$, $\B \neq 0$ and in any frame $\E \cdot \B = 0$, i.e. $\E$ and $\B$ are perpendicular.
\end{itemize}

\section{Energy and Momentum Tensor}

We showed in Chapter 1 that the energy in the fields $w = \frac{1}{2} (E^2 + B^2)$ (a scalar with 1 component), the Poynting vector $$\vec S = c \E \times \B$$ (a vector with 3 components) and the stress tensor $$t_{ij} = E_i E_j + B_i B_j - \frac{1}{2} \delta_{ij} (E^2 + B^2)$$ (a symmetric tensor with 6 independent components). We expect to be able to combine all of these objects into a single symmetric tensor $T_{\mu \nu}$ with $1 + 3 + 6 = 10$ independent components.

$T_{\mu \nu}$ has the following properties
\begin{itemize}
\item Quadratic in $\E$ and $\B$ (i.e. in $F_{\mu \nu}$ and $F^\star_{\mu \nu}$)
\item Symmetric under duality symmetry $\E \rightarrow \B$, $\B \rightarrow - \E$, and the Poynting vector $\vec S \rightarrow \vec S$  (i.e. under $F_{\mu \nu} \rightarrow F^\star_{\mu \nu}$).
\end{itemize}

So consider the ansatz $$T_{\mu \nu} = \frac{1}{2} \left( F_{\mu \alpha} F^\alpha \, _\nu + F^\star_{\mu \alpha} F^{\star \alpha} \, _\nu \right),$$ we can check that this satisfies the above properties
$$F_{0\alpha} F^\alpha \, _0 = F_{0i} F^i \, _0 = E^2$$
$$F_{0 \alpha} F^\alpha \, _j = F_{0i} F^i \, _j = -E_i \epsilon_{ijk} B_k = (\E \times \B)_j$$
\begin{eqnarray*}
F_{i \alpha} F^\alpha \, _j &=& F_{i0} F^0 \, _j + F_{ik} F^k \, _j \\
&=& - E_i E_j - \epsilon_{ijk} \epsilon_{kjm} B_l B_m \\
&=& - E_i E_j + (\delta_{il} \delta_{lm} - \delta_{im} \delta_{jl} ) B_l B_m = -E_i E_j - B_i B_j + \delta_{ij} B^2
\end{eqnarray*}

So

$$T_{00} = \frac{1}{2} (E^2 + B^2)$$
$$T_{0j} = (\E \times \B)_j = \frac{1}{c} S_j$$
$$T_{ij} = - \left(E_i E_j + B_i B_j - \frac{1}{2} \delta_{ij} (E^2 + B^2) \right) = - t_{ij}$$

We could also write $$T_{\mu \nu} = F_{\mu \alpha} F^\alpha \, _\nu + \frac{1}{4} \eta_{\mu \nu} F_{\alpha \beta} F^{\alpha \beta}$$ where $\eta_{00} = 1$, $\eta_{ij} = - \delta_{ij}$. By writing out the components we can see that these expressions are equivalent.

The stress-energy tensor is symmetric, and traceless: $$T_\mu \, ^\mu = - F_{\mu \alpha} F^{\mu \alpha} + \frac{1}{4} \underbrace{ \delta_\mu ^\mu }_4 F_{\alpha \beta} F^{\alpha \beta} = 0.$$

To understand energy-momentum conservation covariantly, consider its divergence (c.f. $\partial^\mu J_\mu = 0$ for the conservation of 4-current)
 \begin{eqnarray*}
 	\partial^\mu T_{\mu \nu} 	&=& \partial^\mu ( F_{\mu \alpha} F^\alpha \, _\nu ) + \frac{1}{4} \partial_\nu ( F_{\alpha \beta} F^{\alpha \beta} ) \\
						&=& (\partial^\mu F_{\mu \alpha} ) F^\alpha + F_{\mu \alpha} \partial^\mu F^\alpha \, _\nu + \frac{1}{2} F_{\alpha \beta} \partial_\nu F^{\alpha \beta} \\
						&=& \underbrace{ \frac{1}{c} J_\alpha F^\alpha \, _\nu }_\textrm{Maxwell} + F^{\alpha \beta}
			%				( \underbrace{ \partial_\alpha F_{\beta \nu} + \frac{1}{2} \partial_\nu F_{\alpha \beta} }_{\frac{1}{2} \partial_\alpha F_{\beta \nu} +
			%				\frac{1}{2} (\partial_\alpha F_{\beta \nu} + \partial_\nu F_{\alpha \beta} ) = \frac{1}{2} ( \partial_\alpha F_{\beta \nu} + \partial_\beta F_{\alpha \nu} )
\end{eqnarray*} Now ${1}{2} ( \partial_\alpha F_{\beta \nu} + \partial_\beta F_{\alpha \nu} )$ is symmetric, but $F^{\alpha \beta}$ is antisymmetric, so %$F^{\alpha \beta} 1}{2} ( \partial_\alpha F_{\beta \nu} + \partial_\beta F_{\alpha \nu} ) = 0$, i.e. the last term vanishes, and we have $$\partial^\mu T_{\mu \nu} = \frac{1}{c} J^\mu F_{\mu \nu}.$$

In the absence of charges and currents $$\partial^\mu = T_{\mu \nu} = 0$$ and we can obtain a continuity equation as we did for charge conservation $$0 = \int_{V_4} \partial^\mu T_{\mu \nu} \d^4 x = \oint_{S_3} \d \Sigma^\mu T_{\mu \nu}.$$ Taking $V_4$ to be a cylinder, as before for charge conservation, we have \emph{four} conserved quantities given by $T_{0 \nu}$

\begin{itemize}
\item $\nu = 0$ $$\frac{1}{c} \diff{}{t} \int_{V_3} \d^3 \vec x \underbrace{ T_{00} (\vec x,t) }_\textrm{Field energy, $u$} = - \oint_{S_2} \d A_i  \underbrace{ T_{i0}(\vec x, t) }_\textrm{Poynting vector $S_i/c$ (energy flow)} $$ i.e. energy is conserved.

\item $\nu = i$ $$\frac{1}{c} \diff{}{t} \int_{V_3} \d^3 \vec x \underbrace{ T_{0i}(\vec x, t) }_\textrm{momentum density $S_i / c$} = - \oint \d A_j \underbrace{ T_{ij} (\vec x, t) }_\textrm{Stress tensor $-t_{ij}$}$$ i.e. momentum is conserved.
\end{itemize}

In the presence of charges (i.e. $J_\mu \neq 0$) $$\oint_{S_3} \d \Sigma^\mu T_{\mu \nu} = \frac{1}{c} \int_{V_4} J^\mu F_{\mu \nu} \d^4 x$$ so we get an extra term on the right hand side.

\begin{itemize}
\item $\nu = 0$ $$\diff{}{t} \int \d^3 \vec x u = - \oint_{S_2} \d A_i S_i + \int \d^3 \vec x \underbrace{ J^i F_{i0} }_\textrm{$-\E \cdot \J$, power density} $$
\item $\nu = i$ $$\diff{}{t} \int \d^x \vec x \frac{1}{c^2} S_i = \oint_{S_2} \d A_j t_{ij} + \int \d^3 \vec x \underbrace{ ( J^0 F_{0i} + J^j F_{ji} ) }_{-(\rho \E + \frac{1}{c} \J \times \B) }$$ i.e. the Lorentz force.
\end{itemize}

So $f_\mu = \frac{1}{c} F_{\mu \nu} J^\nu$ is the Lorentz (4)-force. $f_0 = \E \cdot \J$ gives the rate of change of energy. $\vec f = \rho \E + \J \times \B$ gives the rate of change of momentum. Note $\partial^\mu T_{\mu \nu} = - f_\nu$ is the back reaction of the e.m. force on \emph{fields}.

\subsection{Relativistic Massive Charged Particle}

$J_\mu = q u_\mu$, so the Lorentz force is $\frac{q}{c} F_{\mu \nu} u^\nu$, and the equation of motion is $$m \diff{}{\tau} u_\mu = \frac{q}{c} F_{\mu \nu} u^\nu.$$ Since $u^\mu \diff{}{\tau} u_\mu = 0$, we must have $u^\mu u^\nu F_{\mu \nu} = 0$, but this is guaranteed by antisymmetry of $F_{\mu \nu}$.

In components \begin{eqnarray*} \gamma \diff{}{t} p_i &=& \frac{q}{c} ( F_{i0} u^0 + F_{ij} u^j ) \\
&=& \frac{q}{c} \gamma ( E_i c + \epsilon_{ijk} B_k u_j ) \\
\Rightarrow \diff{}{t} \vec p = q ( \E + \frac{1}{c} \vec u \times \B ).
\end{eqnarray*}
So the Lorentz force gives rate of change of (relativistic) momentum $\vec p = \gamma m \vec u$. N.B. note that this is $\diff{}{t}$, not $\diff{}{\tau}$: the $\gamma$ cancels.

$$\gamma \diff{}{t} \left( \frac{E}{c} \right) = \frac{q}{c} ( F_{oi} u^i ) = \frac{q}{c} \gamma \vec u \cdot \E$$ i.e. $$\diff{}{t} E = q \vec u \cdot \E$$ i.e. the rate of change of (relativistic) energy $E = \gamma mc^2$ is the rate of working of the Lorentz force.

\subsection{Angular Momentum}

As in the non-relativistic treatment, wherever there is momentum, we can define angular momentum. Consider $$L_{\alpha \beta \gamma} = T_{\alpha \beta} x_\gamma - T_{\alpha \gamma} x_\beta, \; L_{\alpha \beta \gamma} = -L_{\alpha \beta \gamma}.$$ Take the divergence of this $$\partial^\alpha L_{\alpha \beta \gamma} = \cancel{ \partial^\alpha  T_{\alpha \beta} } x_\gamma + T_\gamma \beta - \cancel{ \partial^\alpha T_{\alpha \gamma} } x_\beta - T_{\beta \gamma} = 0$$ because $\partial^\alpha T_{\alpha \beta} = 0$, and $T_{\alpha \beta} = T_{\beta \alpha}$.

So we have  further 6(!) conserved quantities $$L_{0\beta \gamma} = T_{0 \beta} x_\gamma - T_{0 \gamma} x_\beta$$

Clearly $$L_{0ij} = T_{0i} x_j - T_{0j} x_i = \frac{1}{c} ( S_i x_j - S_j x_i )$$ so $$\frac{1}{2} \epsilon_{ijk} L_{0kj} = \frac{1}{c} \vec c \times ( \E \times \B )_i$$, i.e. the angular momentum of the fields.

The other three conserved quantities are $$L_{00i} = T_{00} x_i - T_{0i} x_0 = u x_i - t S_i$$ This is a conservation law for the centre of mass of field energy (see example sheet).

\chapter{}

The Lienard formula for radiation from an accelerating charge is:

$$P = \frac{e^2}{6 \pi c} \gamma^6 ( \dot \beta^2 - (\vec \beta \times \dot{\vec \beta})^2 )$$

We can now examine several special cases:

\begin{enumerate}
\item Non-relativistic limit ($\gamma \rightarrow 1$)
$$\diff{P}{\Omega} = \frac{e^2}{16\pi^2 c} | \vec n \times ( \vec n \times \dot{\vec \beta}) |^2 = \frac{e^2}{16 \pi^2 c} \dot \beta^2 \sin^2 \theta$$
i.e. the usual dipole result: maximal radiation is perpendicular to $\dot{\vec \beta}$. Total power $P$ is obtained by integrating the above
$$P = \frac{e^2}{16 \pi^2 c} \dot \beta^2 \int_0^{2\pi} \d \phi \int_{-1}^{1} \d ( \cos \theta ) ( 1- \cos^2 \theta ) = \frac{e^2}{6 \pi c} \dot \beta^2$$
which recovers the non-relativistic Larmor formula.

Note that $P$ is a Lorentz invariant: under Lorentz transformations
$$\d E \rightarrow \gamma \d E, \d t \rightarrow \gamma \d t, \Rightarrow \diff{E}{t} \rightarrow \diff{E}{t}$$

We can use the Larmor formula and covariance to ``derive'' the Lienard formula. Consider:
$$\dot \beta^2 = \left( \diff{\vec \beta}{t} \right)^2 = \left( \frac{1}{\gamma} \diff{\vec \beta}{\tau} \right)^2 = \left( \frac{1}{\gamma c} \diff{\vec u}{\tau} \right)^2$$
since $\vec \beta \equiv \vec u / c$. But
$$\diff{u_\mu}{\tau} \diff{u^\mu}{\tau} = \left( \diff{}{\tau} \gamma[c,\vec u] \right)^2 = \left[0,\gamma \diff{\vec u}{\tau} \right]^2 = - \left( \gamma \diff{\vec u}{\tau} \right)^2 $$
$$\dot \beta^2 \rightarrow - \frac{1}{c^2} \diff{u_\mu}{\tau} \diff{u^\mu}{\tau}$$
But
$$\diff{u_\mu}{\tau} = \gamma \diff{}{t} \gamma c [1,\vec \beta] = \gamma c [ \dot \gamma , \dot \gamma \vec \beta + \gamma \dot{ \vec \beta } ]$$

Now to find the ordinary time derivative of $\gamma$, we can use the following trick. Since by definition $$\gamma^2 ( 1 - \beta^2 ) = 1,$$ taking the time derivative of each side gives $$\gamma \dot \gamma ( 1 - \beta^2 ) - \gamma^2 \vec \beta \cdot \dot{\vec \beta} = 0$$ then by rearranging we obtain $$\dot \gamma = \gamma^3 \vec \beta \cdot \dot{\vec \beta}$$

So
\begin{eqnarray*}
\diff{u_\mu}{\tau} \diff{u^\mu}{\tau} &=& \gamma^2 c^2 ( \dot \gamma^2 - ( \dot \gamma^2 \beta^2 + 2 \gamma \dot \gamma \vec \beta \cdot \dot{\vec \beta} + \gamma^2 \dot \beta^2 )) \\
&=& \gamma^2 c^2 ( \gamma^4 ( \vec \beta \cdot \dot{\vec \beta} )^2 - 2 \gamma^4 (\vec \beta \cdot \dot{\vec \beta} )^2 - \gamma^2 \dot \beta^2 ) \\
&=& - c^2 ( \gamma^4 \dot \beta^2 + \gamma^6 ( \vec \beta \cdot \dot{\vec \beta} )^2 ) \\
&=& - c^2 \gamma^6 ( \dot \beta^2 - (\vec \beta \times \dot{\vec \beta} )^2 )
\end{eqnarray*}
where we have used the fact that
$$(\vec \beta \times \dot{\vec \beta})^2 = \beta^2 \dot \beta^2 - (\vec \beta \cdot \dot{\vec \beta} )^2$$
so relativistically we obtain
$$P = \frac{e^2}{6 \pi c} \left( - \frac{1}{c^2} \diff{u^\mu}{\tau} \diff{u_\mu}{\tau} \right) = \frac{e^2 \gamma^6}{6 \pi c} ( \dot \beta^2 - ( \vec \beta \times \dot{\vec \beta} )^2 )$$
i.e. the Lienard formula.

\item Linear acceleration

$\vec \beta$ and $\dot{\vec \beta}$ are parallel, so $\vec \beta \times \dot{\vec \beta} = 0$

$$\diff{P}{\Omega} = \frac{e^2}{16 \pi^2 c} \frac{ | \vec n \times ( \vec n \times \dot{\vec \beta} ) |^2 }{ ( 1 - \vec n \cdot \vec \beta )^5 } = \frac{e^2}{16 \pi^2 c}  \frac{ \dot \beta^2 \sin^2 \theta }{ ( 1- \beta \cos \theta )^5 }$$
So we get the usual $\sin^2 \theta$ pattern, but heavily boosted in the forward ($\theta \approx 0$) direction. Note that the radiation is in the direction of $\vec \beta$; it tends to slow the particle down (if it were in the other direction, there would be exponential acceleration).

$$P = \frac{e^2}{16 \pi^2 c} \dot \beta^2 \int_0^{2\pi} \d \phi \int_{-1}^1 \d ( \cos \theta ) \frac{ ( 1 - \cos^2 \theta ) }{ ( 1 - \beta \cos \theta )^5 } = \frac{e^2 \dot \beta^2 \gamma^6}{6 \pi c}$$
The $\gamma^6$ factor is a huge enhancement at relativistic velocities. We have used the integral
$$\int_{-1}^1 \d x \frac{1-x^2}{(1 - \beta x)^5} = \frac{4}{3} \frac{1}{(1 - \beta^2)^3}$$

\item Synchrotron Radiation

Consider an electron travelling in a circular orbit with constant angular velocity $\vec \omega$: $\vec u = \vec \omega \times \r$, $\dot{\vec u} = \omega \times \dot \r = \vec \omega \times \vec u$, i.e. the acceleration is directed radially inwards: $\dot{\vec \beta} = \vec \omega \times \vec \beta$

$$\vec \beta \cdot \vec \omega = \dot{\vec \beta} \cdot \vec \omega = \vec \beta \cdot \dot{\vec \beta} = 0$$

$$\vec n \cdot \vec \beta = \beta \cos \theta$$
$$\vec n \cdot \dot{\vec \beta} = \omega \beta \sin \theta \cos \theta$$

$$| \vec n \times ( ( \vec n - \vec \beta ) \times \dot{\vec \beta} |^2 = \dot \beta^2 ( 1 - \vec n \cdot \vec \beta )^2 - ( 1 - \beta^2 )( \vec n \cdot \dot{\vec \beta} )^2$$

So
$$\diff{P}{\Omega} = \frac{e^2}{16 \pi^2 c} \frac{ \omega^2 \beta^2 }{ ( 1 - \beta \cos \theta )^3 } \left( 1 - \frac{ ( 1-\beta^2 ) }{ ( 1 - \beta \cos \theta )^2 } \sin^2 \theta \cos^2 \phi \right)$$
Again when $\beta \rightarrow 1$, $\theta \approx 0$ dominates, i.e. radiation is emitted in a cone, tangential to the orbit
$$P = \frac{e^2 \omega^2 \beta^2}{16 \pi^2 c } 2 \pi \int_{-1}^1 \d ( \cos \theta ) \frac{ 1 }{ (1-\beta \cos \theta)^3 } \left( 1 - \frac{ (1-\beta^2)(1-\cos^2 \theta) }{ (1 - \beta \cos theta)^2 } \frac{1}{2} \right) = \frac{e^2 \omega^2 \beta^2 \gamma^4 }{ 6 \pi c }$$
having used the integral
$$\int_{-1}^1 \d x \frac{1}{(1- \beta x)^3} = \frac{2}{(1 - \beta )^2 }$$

Note the energy of the electron is $E = \gamma m c^2$, and $\beta = \omega r / c$, so for $\beta \rightarrow 1$

$$P \approx \frac{e^2 c}{6 \pi r^2} \left( \frac{E}{mc^2} \right)^4$$

So the power radiated grows very fast with $E/m$ (particularly for $m$ small, e.g. an electron). To keep energy losses down, we need $r$ to be as big as possible (consider LEP at CERN, the largest electron collider ever built).

In astronomy $P$ can be huge (arrives in pulses).

\end{enumerate}

\subsection{Radiative Damping}

An accelerating electron loses energy in the form of EM radiation; we saw above that $$P = - \frac{e^2}{6 \pi c^3} \diff{u_\mu}{\tau} \diff{u^\mu}{\tau} = - m \bar \tau \dot u_\mu \dot u^\mu$$ where constants are absorbed into the constant $\bar \tau$ with dimensions of time (not to be confused with proper time, $\tau$), and $\cdot \equiv \diff{}{\tau}$ (not an ordinary time derivative). $u_\mu$ is the 4-velocity of the electron $$\bar \tau \equiv \frac{e^2}{6 \pi m c^3} \sim 10^{-23} \mathrm{s}$$

We need to add a term into the equation of motion for the relativistic electron to account for this lost energy $$\left. \diff{E}{\tau} \right|_\textrm{rad} = \gamma \diff{E}{t} = - \gamma P$$ so covariantly since $p_\mu = [E/c,\vec p]$, and $u_\mu = \gamma c [1,\vec \beta]$ $$\left. \diff{p_\mu}{\tau} \right|_\textrm{rad} = - \frac{P u_\mu}{c^2}$$

Thus the equation of motion for the electron is $$\diff{p_\mu}{\tau} = \frac{e}{c} F_{\mu \nu} u^\nu - \frac{ P u_\mu}{c^2} + \kappa_\mu$$ where $F_{\mu \nu}$ is the external field, and $\kappa_\mu$ is required to satisfy the consistency condition on $u_\mu$ i.e. that $$u_\mu u^\mu = c^2 \Rightarrow u_\mu \dot u^\mu = 0$$ (in general the consistency condition allows the derivatives of $u_\mu$ to be expressed as lower derivatives e.g. $u_\mu \ddot u^\mu + \dot u_\mu \dot u^\mu = 0$) so $$u^\mu \left( \kappa_\mu - \frac{P}{c^2} u_\mu \right) = 0$$ i.e. $$u^\mu \kappa_\mu = P = - m \bar \tau \dot u_\alpha \dot u^\alpha = m \bar \tau u_\alpha \ddot u^\alpha $$ This must be true for all $u_\mu$, so $\kappa_\mu = m \bar \tau \ddot u_\mu$ is fixed. $\kappa_\mu$ is referred to as the ``Schott term'' (1912).

The equation of motion is thus $$m \dot u_\mu = \frac{e}{c} F_{\mu \nu} u^\nu - \frac{P}{c^2} u_\mu + m \bar \tau \ddot u_\mu$$ the first two terms are the Lorentz force, and back reaction, and they are even under $t \rightarrow -t$, and the right two term are odd under the same transformation. This equation is referred to as the ``Abraham-Lorentz-Dirac'' equation (1934).

Historical note: in the non-relativistic limit (i.e. $u \ll c$) we obtain $$m \dot{\vec u} = e ( \E + \frac{1}{c} \vec u \times \B ) + m \bar \tau \ddot{\vec u}$$ the ``Abraham-Lorentz'' equation (1903-4). Remarkably, Abraham-Lorentz was not derived from Larmor, but from a model of the electron as a spherical shell of charge, of radius $a$. They then considered the force of the electron on itself $$\vec F_\textrm{self} = \frac{2}{3} m \frac{\bar \tau c}{a} \dot{\vec u} + m \bar \tau \ddot{\vec u} + \mathcal{O}(a)$$ (first term is the divergent contribution to mass).

What are the consequences of $\kappa_\mu$?

Rewrite the ALD equation as $$ m ( \dot u_\mu - \bar \tau \ddot u_\mu ) = F_\mu (\tau)$$ where $F_\mu (\tau)$ absorbs the Lorentz and back reaction terms. Because this equation depends upon $\ddot u_\mu$, it has runaway solutions. $$u_\mu = u_\mu^0 e^{\tau/ \bar \tau}$$ is a solution, but this is a disaster, as in just $\bar \tau \sim 10^{-23} \mathrm{s}$ the electron accelerates away, even in the absence of an external force. Fortunately, these runaway solutions can be eliminated by imposing a boundary condition on the \emph{future}. We insist that as $\tau \rightarrow \infty$, i.e. $F_\mu (\tau) \rightarrow 0$, $\dot u(\tau) \rightarrow 0$. We can then solve the differential equation with an integrating factor; write $\dot u_\mu (\tau) = e^{\tau / \bar \tau} a_\mu (\tau)$, and the equation becomes $$m(a_\mu e^{\tau/\bar \tau} - \bar \tau( \frac{1}{\bar \tau} a_\mu + \dot a_\mu ) e^{\tau/\bar \tau} ) = F_\mu$$ terms cancel to give $$\dot a_\mu = - \frac{1}{m \bar \tau} F_\mu e^{-\tau/\bar \tau}$$ i.e. $$a_\mu ( \tau ) = \frac{1}{m \bar \tau} \int_\tau^\infty \d \tau' F_\mu (\tau') e^{-\tau' / \bar \tau}$$ and $$\dot u_\mu (\tau) = \frac{e^{\tau/\bar \tau}}{m \bar \tau} \int_\tau^\infty \d \tau' F_\mu (\tau') e^{-\tau' / \bar \tau}$$

This eliminates the runaway solutions, since $\tau' > \tau$, $e^{(\tau - \tau')/z} < 1$. Let $S=(\tau' - \tau)/\bar \tau$: $$\dot u_\mu (\tau) = \frac{1}{m} \int_0^\infty \d s e^{-s} F_\mu (\tau + s \bar \tau)$$ This is fine... but it violates causality: the force acts \emph{before} it is applied. e.g. consider $F_\mu(\tau) = F_\mu^0 \Theta(\tau)$, then $$\int_0^\infty \d s e^{-s} \Theta(\tau + s \bar \tau) = \left\{ \begin{array}{c}1 \textrm{ if } \tau > 0 \\ e^{\tau/\bar \tau} \textrm{ if } \tau < 0 \end{array} \right.$$ this effect is called ``pre-acceleration''.

Note that if we use $$\bar a \equiv \frac{e^2}{4 \pi m c^2}$$, the ``classical electron radius'' ($\bar a \sim 3 \cdot 10^{-15} \textrm{m}$), then $$\bar \tau = \frac{3}{2} \frac{\bar a}{c} \sim \textrm{time taken to travel $\bar a$} \approx 10^{-4} \textrm{atomic diameter}$$.

The resolution is that, for finite-sized models of an electron (e.g. the spherical shell of charge, radius $a$), it can be shown (although it is technically rather difficult), there are no runaway solutions, and no pre-acceleration, provided that $a \approx > \bar a$. This can be understood as follows: the electrostatic energy of the spherical shell is $\sim \frac{\rho^2}{8 \pi a} = \frac{1}{2} \frac{\bar a}{a}mc^2$. In relativity $E=mc^2$ in the rest frame, and so there is a contribution $\sim \frac{1}{2} \frac{\bar a}{a} m$ to the electron mass. But $$m = m_0 + \frac{1}{2} \frac{\bar a}{a} m$$ i.e. the physical mass is the sum of Newtonian ``bare mass'', and the extra term.

Provided that $m_0 > 0$, everything is fine. But if $m_0 < 0$ there are acausal solutions (pre-acceleration). Classical electrodynamics thus cannot consistently describe massive point particles! As Einstein said ``the electron is a stranger in electrodynamics''.

However, the Compton wavelength $\lambda = \frac{\hbar}{mc} = 4 \cdot 10^{-13} \textrm{m}$, $\lambda > \bar a$ by a factor of around 100: Before classical dynamics fails, quantum dynamics becomes necessary. Is this necessary, or is it just a coincidence that quantum mechanics takes over at this point? Could there be some particle for which the classical problems are exposed? No: $$\frac{\bar a}{\lambda} = \frac{e^2}{4 \pi mc^2} \frac{mc}{\hbar} = \frac{e^2}{4 \pi \lambda c \hbar} = \alpha \sim \frac{1}{137}$$ the ``fine structure constant'', which governs the strength of interactions between photons and electrons. $\alpha \ll 1$, so the self-interaction of an electron (emission and re-absorption of photons) is weak; there are fewer loops on the Feynman diagram. If $\alpha$ were larger, then the photon would be covered by a ``fuzz'' of electrons, and we would no longer be able to observe the electron directly (c.f. quarks and gluons, which are strongly interacting $\alpha_s \sim 1$, are not detected in isolation, but only in bound states i.e. mesons and baryons). The observed quanta would themselves be larger, and again the classical problems would not be detectable.

\chapter{Action Principles}

The formulation of relativistic dynamics using action principles. In Lagrangian dynamics using action makes things easier that using equations of motion, and the same is true for relativistic dynamics, and in quantum theory. Actions form a useful starting point for quantising field theories, as actions make invariances more explicit. Electromagnetism is the paradigm for a field theory, and so it is useful to explore action principles in electromagnetism.

\section{Non-Relativistic Point Particle}

For non-relativistic particles, we consider a path $q(t)$, and the action is $$S[q(t)] = \int_a^b \d t \left( \frac{1}{2} m \dot q^2 - V(q) \right)$$ which is invariant under time translations $t \rightarrow t+t_0$. In non-relativistic dynamics time and space are treated separately (time is a parameter).

Consider a small change in the path $q \rightarrow q + \delta q$, the action changes by $$\delta S = \int_a^b \left( m \dot q \delta \dot q - V' (q) \delta q \right) = - \int_a^b \left( m \ddot q + V'(q) \right) \delta q + \underbrace{ \int_a^b \d t \diff{}{t} ( m \dot q \delta q ) }_{\left. m\dot q \delta q \right|_a^b}$$ where we have integrated by parts. Now Hamilton's Principle requires that the action be stationary i.e. $\delta S = 0$, and since $\delta q(a) = \delta q(b) = 0$, then $$m \ddot q + V'(q) = 0.$$ The (generalised) momentum $$p = \pdiff{\mathcal L}{\dot q} = m \dot q.$$

\section{Relativistic Point Particles}

Now we want $S$ to be invariant under translations $x_\mu \rightarrow x_\mu + a_\mu$, and Lorentz transforms $x_\mu \rightarrow \Lambda_\mu \, ^\nu x_\nu$.

But for a point particle on a path $r_\mu (\tau)$, the only invariants are $m$ and proper time $\tau$, so the only invariant action we can build with the correct units is $$S = - mc^2 \int_a^b \d \tau$$ up to a constant, where dimensional analysis requires $[S] = [Et]$.

Note that $$\d \tau = \gamma^{-1} \d t = \left( 1 - \frac{u^2}{c^2} \right)^{1/2} \d t \approx \left( 1 - \frac{1}{2} \frac{u^2}{c^2} + \cdots \right) \d t$$ if $u \ll c$, so in the nonrelativistic limit $$S \approx - m c^2 \int_a^b \d t \left( 1 - \frac{1}{2} \frac{u^2}{c^2} + \cdots \right) = \int_a^b \d t \left( - mc^ + \frac{1}{2} mu^2 + \cdots \right)$$ i.e. it reduces to the nonrelativistic action (up to a constant, which leaves the equations of motion unchanged).

But there is no integrand! So, how do we vary this action? To find $\delta S$, note $c^2 \d \tau^2 = \d r^\mu \d r_\mu$, so we can write $$\d \tau = \frac{1}{c} \left( \underbrace{ \diff{r_\mu}{\tau} }_{\dot r_\mu(\tau)} \diff{r^\mu}{\tau} \right)^{1/2} \d \tau,$$ then $$S[r_\mu(\tau)] = - mc \int_a^b ( \dot r_\mu \dot r^\mu )^{1/2} \d \tau$$

We only have $\dot r_\mu \dot r^\mu = u_\mu u^\mu = c^2$ \emph{on} the nonrelativistic path; in order to vary $S$ we must go off the nonrelativistic path. $$\delta S = - mc \int_a^b \d \tau \frac{1}{(\dot r_\alpha \dot r^\alpha )^{1/2}} \dot r_\mu \delta \dot r^\mu = - m \int_a^b \d \tau \dot r_\mu \delta r^\mu$$ $$=m \int_a^b \d \tau \delta r^\mu \diff{}{\tau} \dot r_\mu - m \int_a^b \d \tau \diff{}{\tau} ( \underbrace{ \dot r_\mu \delta r^\mu ) }_{\left. \dot r_\mu \delta r^\mu \right|_a^b}$$

So again if $\delta S = 0$, $\delta r_\mu(a) = \delta r_\mu(b) = 0$ $$m \diff{}{\tau} \dot r_\mu = \diff{p_\mu}{\tau} = 0$$ as expected. Canonical momentum $$p_\mu = \pdiff{\mathcal L}{\dot r_\mu}.$$

\section{Electromagnetic Fields}

Again we need an $S$ which is invariant under space-time translations and rotations. By encoding the symmetries in the action, we are guaranteed to find equations of motion which are covariant. It is generally much easier to think of things which are \emph{invariant} (actions) than to think of things which are \emph{covariant}.

$F_{\mu \nu} (x)$ is a \emph{field}. To find an invariant we need to integrate over all $x_\mu$. Now $\d^4 x$ is an invariant integration measure, and $F_{\mu \nu} F^{\mu \nu}$, $F^\star_{\mu \nu} F^{\mu \nu}$ are Lorentz invariant, as is $A_\mu J^\mu$ (for some fixed current density $J_\mu(x)$.

%$$[\d^4 x] = [L^4], \; [A] = [QL^{-1}], \; [E} = [QL^{-2}]$$

Try $$S[A_\mu] = \frac{1}{c} \underbrace{ \int_V \d^4 x }_{c \int \d t \int \d^3 \vec x} \left( \underbrace{ \frac{1}{4} F_{\mu \nu} F^{\mu \nu} + \frac{1}{c} J_\mu A^\mu }_\textrm{Lagrangian density} \right)$$

Much like solving a differential equation, finding the appropriate Lagrangian involves making a guess, and checking that it reproduces the known solution (i.e. the correct equations of motion). Hence we vary this Lagrangian $$\delta S = \frac{1}{c} \int \d^4 x \left( \frac{1}{2} F_{\mu \nu} \delta F^{\mu \nu} + \frac{1}{c} J_\mu \delta A^\mu \right)$$
$$= \frac{1}{c} \int \d^4 x \left( \frac{1}{2} \underbrace{ F_{\mu \nu} ( \partial^\mu \delta A^\nu - \partial^\nu \delta A_\mu ) }_{F_{\mu \nu} \partial^\mu \delta A^\nu} + \frac{1}{c} J_\mu \delta A^\mu \right)$$
$$= \frac{1}{c} \int_V \d^4 x \left( - \partial^\mu F_{\mu \nu} + \frac{1}{c} J_\nu \right) \delta A^\nu + \frac{1}{c^2} \int_S \d \Sigma^\mu F_{\mu \nu} \delta A^\nu$$ using the 4-dimensional divergence theorem.

So for variations $\delta A_\mu$ such that $\delta A_\mu = 0$ on the boundary of the surface $S$ $$\delta S = 0 \Rightarrow \partial^\mu F_{\mu \nu} = \frac{1}{c} J_{\nu}$$ i.e. the Maxwell equations.

Maxwell's equations are invariant of the Lorentz transformations, and also invariant under gauge transformations. Note that although $F_{\mu \nu}$ is invariant under the gauge transformation $A_\mu \rightarrow A_\mu + \partial_\mu \chi$, but the right-hand term in the Lagrangian density $J_\mu A^\mu$ is \emph{not} invariant under the gauge transformation. However we are concerned with the integral of this term i.e. $$\int d^4 x A_\mu J^\mu \rightarrow \int d^4 x \left( A^\mu J_\mu + \partial^\mu \chi J_\mu \right)$$ using integration by parts this is equal to $$\int_V \left( A_\mu J^\mu - \chi \partial^\mu J_\mu - \partial^\mu (\chi A_\mu) \right) = \int_V \d^4 x \left( A^\mu J_mu + \int \d \Sigma^\mu \chi A_\mu \right)$$ but the $\chi \partial^\mu J_\mu$ term must vanish to ensure consistency (current conservation). The surface term makes no contribution to the equation of motion, as $A_\mu$ must be fixed on the bounding surface.

So we have gauge invariance of the equations of motion only if $\partial^\mu J_\mu = 0$ i.e. if current is conserved.

Note that the conjugate momenta $$\pi_\mu = \pdiff{\mathcal L}{\dot A_\mu} = - F_{0 \mu} = \left\{ 0 \textrm{ if } \mu = 0 \\ E_i \textrm{ if } \mu = i \right.$$ so one of the conjugate momenta vanishes! Because of gauge invariance only 3 of the 4 components of $A_\mu$ are dynamical.

We can see this directly from Maxwell's equations $$\partial^\mu F_{\mu \nu} = \frac{1}{c} J_\nu,$$ i.e. $$\partial^2 A_\nu - \partial^\mu \partial_\nu A_\mu = \frac{1}{c} J_\nu$$ Taking the Fourier transform $$A_\mu(x) = \int \d^4 k \frac{1}{(2 \pi)^4} e^{i k \cdot x} \tilde A_\mu (k)$$ where $k \cdot x \equiv k_\mu x^\mu$ $$k^2 \tilde A_\nu - k^\mu k_\nu \tilde A_\mu = - \frac{1}{c} \tilde J_\nu$$ i.e. $$(k^2 \eta_{\mu \nu} - k_\mu k_\nu ) \tilde A^\nu = - \frac{1}{c} \tilde J^\nu$$ To solve this we need to find the inverse of $K^2 \eta_{\mu \nu} - k_\mu k_\nu \equiv \Delta_{\mu \nu}$. But $\Delta_{\mu \nu} k^\nu = k^2 k_\mu - k_\mu k^2 = 0$ i.e. $k^\nu$ is a zero eigenvector: $\Delta_{\mu \nu}$ has no inverse! So we \emph{must} fix a gauge to be able to solve this equation.

\section{Gauge Fixing}

So we concluded that gauge fixing isn't ust something you do to make things easier, but in fact, without a fixed gauge the equations of motion are insoluble. Somehow, we have to fix the gauge. The simplest choice is the Lorentz gauge.

$$\partial^\mu A_\mu = 0 ( \textrm{i.e. } k^\mu \tilde A_\mu = 0 )$$

We can do this in the action formulation by introducing a Lagrange multipliers (actually, a field of Lagrange multipliers) $B(x0$. We write the action as

$$\frac{1}{c} \int_{V_4} \d^4 x \left( \frac{1}{4} F_{\mu \nu} F^{\mu \nu} + B(x) \partial^\mu A_\mu + \frac{1}{2} \xi B^2 + \frac{1}{c} J_\mu A^\mu \right)	$$

where $\xi$ is an arbitrary (real) parameter: the `gauge parameter'. If we assume Hamilton's Principle, the EOMs that we obtain are those from varying $A_\mu$ and varying $B$, then

$$\delta S = 0 \Rightarrow \partial^\mu F_{\mu \nu} = \frac{1}{c} J_\nu + \partial_\nu B \; (\delta A_\mu)$$

CHECK SIGNS! SHOULD THERE BE A - SIGN?

$$\partial^\mu A_\mu = - \xi B \; (\delta B)$$

Notice that since the current is conserved, if we take the divergence of the above equation $\partial^\mu \delta A_\mu$, since $\partial ^\mu J_\mu = 0$ (current conservation) then

$$\partial^2 B = 0 \; : \; B \textrm{ is a free field}$$

$$\partial^2 A_\nu - \partial^\mu \partial_\nu A_\mu = - \frac{1}{\xi} \partial_\nu \partial^\mu A_\mu + \frac{1}{c} J_\nu$$

i.e.

$$\left( \eta_{\mu \nu} \partial^2 - (1 - \frac{1}{\xi}) \partial_\mu \partial_\nu \right) A^\nu = \frac{1}{c} J_\mu$$

or in momentum space

$$\left( \underbrace{ \eta_{\mu \nu} K^2 - (1-\frac{1}{\xi}) k_\mu k_\nu }_{\Delta^\xi_{\mu \nu}} \right) \tilde A^\nu = - \frac{1}{c} \tilde J_\mu$$

This is now invertible:

$$\left( \Delta^\xi_{\mu \nu} \right)^{-1} = \frac{1}{k^2} \left( \eta_{\mu \nu} - (1-\xi) k_\mu k_\nu / k^2 \right)$$

we have now fixed the gauge, up to the parameter $\xi$. This should not affect the physics, but can be useful for calculations. If $\xi$ is carried along, then terms involving $\xi$ should cancel at the end of the calculation. Otherwise, we may choose the most obvious $\xi = 1$, called ``Feynman gauge''.

\section{Relativistic Interacting Point Particle}

In the Maxwell action, we already had a term which coupled the external field to the current

$$\frac{1}{c} \int \d^4 x \frac{1}{c} J_\mu A^\mu$$

For a point particle, we have the Lienart-Weichert current (integral along the wordline of the particle)

$$J_\mu(x) = ec\int \d \tau a_\mu \delta^4 \left( x\mu - r_\mu (\tau) \right)$$

If we substitute the above into the maxwell action, it tells us how the point particle couples to the action, then

$$\frac{1}{c} \int \d^4 x e \int \d \tau u_\mu(\tau) \delta^4 ( x_\mu - r_\mu (\tau) ) A^\mu(x) = \frac{e}{c} \int \d \tau u_\mu (\tau) A^\mu \left( r(\tau) \right)$$

so the action for a massive pont particle in an em field is

$$S[r(\tau)] = -mc \int \d \tau ( \dot r_\mu \dot r^\mu )^{1/2} + \frac{e}{c} \int \d \tau \dot r_\mu (\tau) A^\mu \left( r(\tau) \right)$$

So to get the equations of motion we let $r(\tau) \rightarrow r + \delta r$ in the usual way

$$\delta S = - \int \d \tau \delta r^\mu m \ddot r_\mu + \frac{e}{c} \int \d \tau \left( \underbrace{ \diff{}{\tau} \delta r_\mu (\tau) A^\mu \left( r(\tau) \right) }_{ - \delta r_\mu (\tau) \diff{}{\tau} A^\mu \left( r(\tau) \right) + \diff{}{\tau} \left( \delta r_\mu A^\mu \right) } + \dot r_\mu (\tau) \partial_\nu A^\mu \delta r_\nu (\tau) \right)$$

where the left hand term is as before, the same piece as the relativistic point particle on its own.

$$\delta S = - \int \d \tau \delta r^\mu \left( m \ddot r_\mu + \frac{e}{c} ( \underbrace{ \partial_\nu A_\mu \dot r^\nu - \partial_\mu A_\nu \dot r ^\nu }_{-F_{\mu \nu} \dot r^\nu} ) \right)$$

$\delta r_\mu = 0$ at $a$ and $b$

So

$$\delta S = 0 \Rightarrow m \ddot r_\mu = \frac{e}{c} F_{\mu \nu} \dot r^\nu$$

i.e.

$$m \diff{u_\mu}{\tau} = \underbrace{ \frac{e}{c} F_{\mu \nu} u^\nu }_\textrm{Lorentz force}$$

\section{Interacting Paticles}

Consider two point particles with worldlines $r_\mu (\tau)$, $r'_\mu (\tau')$.

One particle will produce a field, which will have an effect on the other particle. If we consdier the effect of $r'$ on $r$, then

$$\frac{e}{c} \int \d \tau \dot r^\mu (\tau) \underbrace{ A'_\mu \left( r(\tau) \right) }_\textrm{field due to $r'$, experienced at $r$} = \frac{e}{c} \int \d \tau \dot r^\mu (\tau) \frac{e}{2 \pi} \int \d \tau' \Theta(r_0 (\tau) - r'_0 (\tau') ) \delta \left( ( (r_\mu(\tau) - r'_\mu (\tau') )^2 \right) \dot r_\mu' (\tau')$$

since the field due to $r'$ is just the Lienard-Weichert potential. But of course there must be a corresponding effect accounting for the effect of $r$ on $r'$

$$\frac{e}{c} \int \d \tau' \dot r'_\mu (\tau') A^\mu \left( r'(\tau') \right) = \frac{e^2}{2 \pi c} \int \d \tau' \d \tau \dot r'_\mu (\tau') \dot r^\mu (\tau) \delta\left( (r' - r)^2 \right) \Theta (r'_0 - r_0)$$

These expressions are symmetric under $r' \leftrightarrow r$, except for the $\Theta$ term. This is as we expect, since there was no precedence of $r'$ over $r$ in the construction. This makes it easy to add the two contributions together, and we obtain something involving

$$\Theta(r_0 - r'_0) + \Theta(r_0 + r'_0) \equiv 1$$

With the labels we have chosen, the effect experienced by $r$ due to $r'$ is at the retarded time, and the effect on $r'$ due to $r$ is at the advanced time. If we exchange labels, this condition is reversed.

So we found altogether

$$S[r_\mu] = -mc \left( \underbrace{ \int \d \tau ( \dot r_\mu \dot r^\mu )^{1/2} + \int \d \tau' (\dot r'_\mu \dot r'^\mu)^{1/2} }_\textrm{kinetic terms} \right) + { \frac{e^2}{4 \pi} }_\textrm{factor of $1/2$: no double counting} \underbrace{ \int \d \tau \int \d \tau' \dot r_\mu (\tau) \dot r'^\mu (\tau') \delta \left( (r-r')^2 \right) }_\textrm{interaction}$$

this is the ``Fokker action'' (1929).

Note that the interaction takes place through the Green's function

$$\frac{1}{4 \pi} \delta (x^2) = \frac{1}{2} \left( G_R(x) + G_A(x) \right) \equiv G_F(x)$$

the ``Feynman Green's function''. Since there is this symmetry in the interacting particles, the important Green's function is neither the retarded nor the advanced, but rather the average of the two.

Which Green's function to use depends upon the questions you ask. Normally we use $G_R$ because we know about the stimulus and want to know the response. This situation is different however, and depends equally on stimuli now and in the future; the interactions are necessarily symmetrical, so we need $G_F$.

Note that there is no self-force or back reaction explicit in these equations. This is because it has already been taken care of:

$$F_{\mu \nu} ^F = \frac{1}{2} \left( F_{\mu \nu} ^R + F_{\mu \nu} ^A \right) = F_{\mu \nu}^R - \frac{1}{2} \left( \underbrace{ F_{\mu \nu}^R - F_{\mu \nu}^A }_\textrm{odd terms under $t \leftrightarrow t'$} \right)$$

It can be shown using the equation of motion that the second term is the radiation reaction (Dirac, 1938).

So the full action for classical electromagnetism

$$S[r_\mu, A_\mu] = { - mc \sum \int \d \tau \left( \dot r_\mu \dot r^\mu \right)^{1/2} }_\textrm{electrons} + { \frac{1}{c} \int \d^4 x \frac{1}{4} F_{\mu \nu} F^{\mu \nu} }_\textrm{electromagnetic fields} + { \frac{e}{c} \sum \int \d \tau \dot r_\mu (\tau) A^\mu \left( r(\tau) \right) }_\textrm{interactions}$$

\emph{provided} that we use $G_F$ when solving for the em field.

\center
\huge{Fin.}

\end{document}
